Thinking about ‘Think of a Number’ Tricks
?In-text Question — How would you change this game to make the final answer 3? What about 5?
The game: think of a number \(x\), double it, add 4, divide by 2, subtract the original number.
\(x\to2x\to2x+4\to x+2\to(x+2)-x=2\)
The number added in the second step decides the answer, because the answer is half of it.
Final answer 3: add 6 instead of 4.
\(x\to2x\to2x+6\to x+3\to(x+3)-x=3\)
Final answer 5: add 10 instead of 4.
\(x\to2x\to2x+10\to x+5\to(x+5)-x=5\)
?In-text Question — Can you come up with more complicated steps that always lead to the same final value?
Think of a number \(x\).
Multiply it by 3, add 12, divide by 3, then subtract the original number.
\(x\to3x\to3x+12\to x+4\to(x+4)-x=4\)
Check with \(x=5\): \(5\to15\to27\to9\to9-5=4\) ✓
A longer trick: think of \(x\), multiply by 6, add 18, divide by 3, then subtract twice the original number.
\(x\to6x\to6x+18\to2x+6\to(2x+6)-2x=6\)
?In-text Question — Find the dates if the final answers are the following:
(i) 1269
(ii) 394
(iii) 296
The steps give \(100M+165+D\), where \(M\) is the month and \(D\) is the day.
So subtract 165 from the answer.
The last 2 digits are the day and the digits before them are the month.
(i) 1269
\(1269-165=1104\)
Day \(=04\), month \(=11\).
(ii) 394
\(394-165=229\)
Day \(=29\), month \(=2\).
Note: 29 February exists only in a leap year, so this date must belong to a leap year.
(iii) 296
\(296-165=131\)
Day \(=31\), month \(=1\).
(ii) 29th February (29/02)
(iii) 31st January (31/01)
MTMath Talk — Can you change the steps in this trick and still find the original date? Instead of subtracting 165 from the final answer, you might have to subtract some other number.
Yes.
The month must end up multiplied by 100, so that the day (at most 2 digits) sits in the last two places.
New trick: multiply the month by 50, add 7, multiply by 2, then add the day.
\(50M\to50M+7\to100M+14\to100M+14+D\)
So now subtract 14 from the answer.
The last 2 digits are the day and the digits before them are the month.
Check with 26/01: \(50\to57\to114\to114+26=140\)
\(140-14=126\).
Day \(=26\), month \(=1\) ✓
MTMath Talk — Try to devise your own ‘Think of a Number’ trick.
Think of a number \(x\).
Add 5, multiply by 2, subtract 4, divide by 2, then subtract the original number.
\(x\to x+5\to2x+10\to2x+6\to x+3\to(x+3)-x=3\)
Check with \(x=9\): \(9\to14\to28\to24\to12\to12-9=3\) ✓
?In-text Question — Use the same rule to fill these pyramids:
Each number is the sum of the two numbers directly below it.
Pyramid 1: \(6+2=8\)
Pyramid 2: second row \(3+4=7\) and \(4+3=7\).
Top \(=7+7=14\)
Pyramid 3: third row \(5+4=9\), \(4+5=9\), \(5+0=5\).
Second row \(9+9=18\) and \(9+5=14\).
Top \(=18+14=32\)
?In-text Question — Fill the following pyramids:
For a 4-row pyramid with bottom row \(a,b,c,d\), the top is \(a+3b+3c+d\) (found in Figure it Out, Question 2 below).
Here we fill each pyramid directly.
Pyramid 1 (bottom row \(4,x,6,y\)):
Third row: \(4+x\), \(x+6\), \(6+y\).
Second row, right box: \((x+6)+(6+y)=22\), so \(x+y=10\).
Top: \(4+3x+18+y=50\), so \(3x+y=28\).
Subtracting: \(2x=18\), so \(x=9\) and \(y=1\).
Pyramid 2 (bottom row \(5,x,7,y\)):
Third row, right box: \(7+y=9\), so \(y=2\).
Second row, left box: \((5+x)+(x+7)=40\), so \(2x=28\) and \(x=14\).
Third row: \(19,21,9\).
Second row: \(40,30\).
Top \(=40+30=70\).
Pyramid 3 (bottom row \(3,5,x,y\)):
Third row, right box: \(x+y=7\).
Top: \(3+15+3x+y=35\), so \(3x+y=17\).
Subtracting: \(2x=10\), so \(x=5\) and \(y=2\).
Third row: \(8,10,7\).
Second row: \(18,17\).
Figure it Out — Set 1
Six questions from page 140 of the textbook.
1Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.
For a pyramid with 3 rows and bottom row \(a,b,c\), the second row is \(a+b\) and \(b+c\).
Top \(=(a+b)+(b+c)=a+2b+c\)
Bottom row 4, 13, 8:
\(4+2\times13+8=4+26+8=38\)
Bottom row 7, 11, 3:
\(7+2\times11+3=7+22+3=32\)
Bottom row 10, 14, 25:
\(10+2\times14+25=10+28+25=63\)
2Write an expression for the topmost row of a pyramid with 4 rows in terms of the values in the bottom row.
Let the bottom row be \(a,b,c,d\).
Third row: \(a+b\), \(b+c\), \(c+d\).
Second row: \((a+b)+(b+c)=a+2b+c\) and \((b+c)+(c+d)=b+2c+d\).
Top: \((a+2b+c)+(b+2c+d)=a+3b+3c+d\)
3Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.
From Question 2, for a 4-row pyramid the top is \(a+3b+3c+d\).
Bottom row 8, 19, 21, 13:
\(8+3\times19+3\times21+13=8+57+63+13=141\)
Bottom row 7, 18, 19, 6:
\(7+3\times18+3\times19+6=7+54+57+6=124\)
Bottom row 9, 7, 5, 11:
\(9+3\times7+3\times5+11=9+21+15+11=56\)
4Recall the Virahāṅka-Fibonacci number sequence 1, 2, 3, 5, … where each number is the sum of the two numbers before it.
If the first three Virahāṅka-Fibonacci numbers are written in the bottom row of a number pyramid with three rows, fill in the rest of the pyramid. What numbers appear in the grid? What is the number at the top? Are they all Virahāṅka-Fibonacci numbers?
The first three numbers are \(1,2,3\).
Second row: \(1+2=3\) and \(2+3=5\).
Top: \(3+5=8\)
The numbers in the grid are 1, 2, 3, 3, 5 and 8.
The number at the top is 8.
The sequence is 1, 2, 3, 5, 8, 13, … so 1, 2, 3, 5 and 8 are all Virahāṅka-Fibonacci numbers.
5What can you say about the numbers in the pyramid and the number at the top in the following cases?
(i) The first four Virahāṅka-Fibonacci numbers are written in the bottom row of a four row pyramid.
(ii) The first 29 Virahāṅka-Fibonacci numbers are written in the bottom row of a 29 row pyramid.
(i) Four rows
Bottom row: \(1,2,3,5\).
Third row: \(3,5,8\).
Second row: \(8,13\).
Top: \(21\)
Every number in the pyramid is a Virahāṅka-Fibonacci number.
Each row is a run of consecutive terms of the sequence: \(1,2,3,5\) then \(3,5,8\) then \(8,13\) then \(21\).
The top, 21, is the 7th term of the sequence \(1,2,3,5,8,13,21\).
(ii) 29 rows
Adding two consecutive terms gives the next term, so every row is again a run of consecutive Virahāṅka-Fibonacci terms.
Each row starts 2 terms later than the row below it.
The bottom row starts at the 1st term, the next row at the 3rd, then the 5th, and so on.
The 29th row (the top) starts at the \(2\times29-1=57\)th term.
So every number in the pyramid is a Virahāṅka-Fibonacci number, and the top is the 57th term of the sequence, 591286729879.
(ii) All numbers are Virahāṅka-Fibonacci numbers; the top is the 57th term, 591286729879.
6If the bottom row of an n row pyramid contains the first n Virahāṅka-Fibonacci numbers, what can we say about the numbers in the pyramid? What can we say about the number at the top?
Let \(V_1,V_2,V_3,\dots\) be the Virahāṅka-Fibonacci numbers \(1,2,3,5,\dots\), so \(V_k+V_{k+1}=V_{k+2}\).
The bottom row is \(V_1,V_2,\dots,V_n\).
Each number in the next row is a sum of two neighbours \(V_k+V_{k+1}=V_{k+2}\).
So the second row is \(V_3,V_4,\dots,V_{n+1}\), which is again a run of consecutive terms.
The same happens in every row, and each row starts 2 terms later than the row below it.
Row \(r\) (counting the bottom row as row 1) starts at \(V_{2r-1}\).
The top is row \(n\), so it is \(V_{2n-1}\).
Check: \(n=3\) gives \(V_5=8\), and \(n=4\) gives \(V_7=21\) ✓
Fun with Grids
MTMath Talk — Create your own calendar trick. For instance, choose a grid of a different size and shape.August 2025Numbers 2 to 50 in rows of 9
My trick: choose a 3 × 3 grid from the calendar, add the 9 numbers and tell me the sum.
Let \(a\) be the top left number.
In a calendar, the next day is +1 and the next week is +7.
First row: \(a,\ a+1,\ a+2\)
Second row: \(a+7,\ a+8,\ a+9\)
Third row: \(a+14,\ a+15,\ a+16\)
Sum \(=9a+(1+2+7+8+9+14+15+16)=9a+72\)
Finding the grid from the sum: suppose the sum is 135.
\(9a+72=135\)
\(9a=63\) (subtracting 72 from both sides)
\(a=7\) (dividing both sides by 9)
So the grid has top left number 7.
Rows: \(7,8,9\), then \(14,15,16\), then \(21,22,23\).
Check: \(24+45+66=135\) ✓
Algebra Grids
?In-text Question — In the following grids, find the values of the shapes and fill in the empty squares:
In each row, the last column is the sum of the values to its left.
First grid (square \(=s\), circle \(=c\)):
Row 1: \(s+s+c=27\), so \(2s+c=27\).
Row 2: \(c+c+s=21\), so \(s+2c=21\).
From Row 1, \(c=27-2s\).
Putting this in Row 2: \(s+2(27-2s)=21\)
\(s+54-4s=21\)
\(54-3s=21\), so \(3s=33\) and \(s=11\).
Then \(c=27-22=5\).
Check: \(11+11+5=27\) ✓ and \(5+5+11=21\) ✓
Row 3: \(c+s+c=5+11+5=21\)
Second grid (circle \(=c\), diamond \(=d\)):
Row 1: \(c+d+d=18\), so \(c+2d=18\).
Row 2: \(d+c+c=15\), so \(d+2c=15\).
From Row 2, \(d=15-2c\).
Putting this in Row 1: \(c+2(15-2c)=18\)
\(c+30-4c=18\)
\(30-3c=18\), so \(3c=12\) and \(c=4\).
Then \(d=15-8=7\).
Check: \(4+7+7=18\) ✓ and \(7+4+4=15\) ✓
Row 3: \(d+c+c=7+4+4=15\)
The last row gives the column sums: \(4+7+7=18\), \(7+4+4=15\) and \(7+4+4=15\).
The corner square is \(18+15+15=48\).
Second grid: circle = 4, diamond = 7; empty squares: 15 in the third row, then 18, 15, 15 and 48 in the last row.
Figure it Out — Set 2
Two questions from page 144 of the textbook.
1Fill the digits 1, 3, and 7 in □□ × □ to make the largest product possible.
The largest digit (7) should be the multiplier.
The other two digits are written in decreasing order to form the multiplicand.
So the arrangement is \(31\times7\).
\(31\times7=217\)
Checking all six arrangements:
| Arrangement | Product |
|---|---|
| 37 × 1 | 37 |
| 73 × 1 | 73 |
| 17 × 3 | 51 |
| 71 × 3 | 213 |
| 13 × 7 | 91 |
| 31 × 7 | 217 |
The largest product is 217.
2Fill the digits 3, 5, and 9 in □□ × □ to make the largest product possible.
The largest digit (9) should be the multiplier.
The other two digits are written in decreasing order to form the multiplicand.
So the arrangement is \(53\times9\).
\(53\times9=477\)
Checking all six arrangements:
| Arrangement | Product |
|---|---|
| 59 × 3 | 177 |
| 95 × 3 | 285 |
| 39 × 5 | 195 |
| 93 × 5 | 465 |
| 35 × 9 | 315 |
| 53 × 9 | 477 |
The largest product is 477.
Decoding Divisibility Tricks
?In-text Question — Can you work out what happens if a > b?
The two-digit number is \(ab=10a+b\).
When reversed, it becomes \(ba=10b+a\).
If \(a\gt b\), then \(ab\gt ba\), so the difference is \(ab-ba\).
\((10a+b)-(10b+a)\)
\(=10a-a+b-10b\)
\(=9a-9b=9(a-b)\)
The difference is again divisible by 9.
Check with 74 and 47: \(74-47=27=9\times3\) ✓
Figure it Out — Set 3
Eleven questions from pages 145–147 of the textbook.
1In the trick given above, what is the quotient when you divide by 9? Is there a relationship between the two numbers and the quotient?
The trick: take a 2-digit number \(ab\) with different digits, reverse it to get \(ba\), find the difference and divide by 9.
If \(b\gt a\), the difference is \((10b+a)-(10a+b)=9(b-a)\).
So the quotient is \(\dfrac{9(b-a)}{9}=b-a\).
If \(a\gt b\), the quotient is \(a-b\) in the same way.
The quotient is the difference of the two digits of the number.
Example: \(47\) and \(74\).
\(74-47=27\) and \(27\div9=3=7-4\) ✓
Example: \(82\) and \(28\).
\(82-28=54\) and \(54\div9=6=8-2\) ✓
2In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen? For example:
• We start with 31. After reversing we get 13. Adding 31 and 13, we get 44.
• We start with 28. After reversing we get 82. Adding 28 and 82, we get 110.
• We start with 12. After reversing we get 21. Adding 12 and 21, we get 33.
Observe that all these numbers are divisible by 11. Is this always true? Can we justify this claim using algebra?
Let the number be \(ab=10a+b\).
Reversed, it is \(ba=10b+a\).
Sum \(=(10a+b)+(10b+a)\)
\(=11a+11b\)
\(=11(a+b)\)
The sum is 11 times a whole number, so it is always divisible by 11.
The quotient is \(a+b\), the sum of the digits.
Check: \(31+13=44=11\times4\), and \(28+82=110=11\times10\) ✓
3Consider any 3-digit number, say abc (100a + 10b + c). Make two other 3-digit numbers from these digits by cycling these digits around, yielding bca and cab. Now add the three numbers. Using algebra, justify that the sum is always divisible by 37. Will it also always be divisible by 3? [Hint: Look at some multiples of 37.]
\(abc=100a+10b+c\)
\(bca=100b+10c+a\)
\(cab=100c+10a+b\)
Sum \(=(100a+10a+a)+(100b+10b+b)+(100c+10c+c)\)
\(=111a+111b+111c\)
\(=111(a+b+c)\)
Now \(111=3\times37\), so the sum \(=37\times3(a+b+c)\).
So the sum is always divisible by 37.
It is also always divisible by 3, because \(111=3\times37\) means the sum is \(3\times37\times(a+b+c)\).
The multiples of 37 are \(37,74,111,\dots\), and 111 is the first one divisible by 3.
Check with 123: \(123+231+312=666=111\times6\) ✓
4Math Talk — Consider any 3-digit number, say abc. Make it a 6-digit number by repeating the digits, that is abcabc. Divide this number by 7, then by 11, and finally by 13. What do you get? Try this with other numbers. Figure out why it works. [Hint: Multiply 7, 11 and 13.]
Take \(abc=234\).
Then \(abcabc=234234\).
\(234234\div7=33462\)
\(33462\div11=3042\)
\(3042\div13=234\)
We get back the original 3-digit number.
Why it works
\(abcabc=1000\times abc+abc\)
\(=1001\times abc\)
Now \(7\times11\times13=77\times13=1001\).
So \(abcabc=7\times11\times13\times abc\).
Dividing by 7, then 11, then 13 leaves \(abc\).
5There are 3 shrines, each with a magical pond in the front. If anyone dips flowers into these magical ponds, the number of flowers doubles. A person has some flowers. He dips them all in the first pond and then places some flowers in shrine 1. Next, he dips the remaining flowers in the second pond and places some flowers in shrine 2. Finally, he dips the remaining flowers in the third pond and then places them all in shrine 3. If he placed an equal number of flowers in each shrine, how many flowers did he start with? How many flowers did he place in each shrine?
Suppose he starts with \(x\) flowers and places \(k\) flowers in each shrine.
After pond 1: \(2x\).
After placing \(k\) in shrine 1: \(2x-k\).
After pond 2: \(2(2x-k)=4x-2k\).
After placing \(k\) in shrine 2: \(4x-3k\).
After pond 3: \(2(4x-3k)=8x-6k\).
All of these go into shrine 3, so \(8x-6k=k\).
\(8x=7k\)
So \(x=\dfrac{7k}{8}\).
For whole numbers, the smallest choice is \(k=8\), which gives \(x=7\).
Check: \(7\to14\), place 8, left 6.
\(6\to12\), place 8, left 4.
\(4\to8\), place 8 ✓
(In general, \(x=7m\) and \(k=8m\); the smallest case is 7 and 8.)
6Math Talk — A farm has some horses and hens. The total number of heads of these animals is 55 and the total number of legs is 150. How many horses and how many hens are on the farm? Can you solve this without letter-numbers? [Hint: If all the 55 animals were hens, then how many legs would there be? Using the difference between this number and 150, can you find the number of horses?]
Using algebra
Let the number of horses be \(h\).
Then the number of hens is \(55-h\).
Legs: \(4h+2(55-h)=150\)
\(4h+110-2h=150\)
\(2h=40\), so \(h=20\).
Hens \(=55-20=35\)
Without letter-numbers
If all 55 animals were hens, the legs would be \(55\times2=110\).
The actual number of legs is 150, so there are \(150-110=40\) extra legs.
Each horse has 2 more legs than a hen, so the number of horses is \(40\div2=20\).
Hens \(=55-20=35\)
Check: \(20\times4+35\times2=80+70=150\) ✓
7A mother is 5 times her daughter’s age. In 6 years’ time, the mother will be 3 times her daughter’s age. How old is the daughter now?
Let the daughter’s age now be \(d\) years.
Then the mother’s age is \(5d\) years.
After 6 years, the daughter is \(d+6\) and the mother is \(5d+6\).
\(5d+6=3(d+6)\)
\(5d+6=3d+18\)
\(2d=12\) (subtracting \(3d\) and 6 from both sides)
\(d=6\)
Check: the mother is \(30\).
After 6 years: \(36=3\times12\) ✓
8Two friends, Gauri and Naina, are cowherds. One day, they pass each other on the road with their cows. Gauri says to Naina, “You have twice as many cows as I do”. Naina says, “That’s true, but if I gave you three of my cows, we would each have the same number of cows”. How many cows do Gauri and Naina have?
Let Gauri have \(g\) cows.
Then Naina has \(2g\) cows.
If Naina gives 3 cows to Gauri, Gauri has \(g+3\) and Naina has \(2g-3\).
These are equal: \(g+3=2g-3\).
\(6=g\) (adding 3 and subtracting \(g\) on both sides)
So \(g=6\) and Naina has \(2g=12\).
Check: \(6+3=9\) and \(12-3=9\) ✓
9I run a small dosa cart and my expenses are as follows:
• Rent for the dosa cart is ₹5000 per day.
• The cost of making one dosa (including all the ingredients and fuel) is ₹10.
(i) If I can sell 100 dosas a day, what should be the selling price of my dosa to make a profit of ₹2000?
(ii) If my customers are willing to pay only ₹50 for a dosa, how many dosas should I aim to sell in a day to make a profit of ₹2000?
Profit \(=\) Income \(-\) Expenses.
(i) 100 dosas a day
Let the selling price of a dosa be ₹\(p\).
Income \(=100p\)
Expenses \(=5000+100\times10=5000+1000=6000\)
Profit: \(100p-6000=2000\)
\(100p=8000\)
\(p=80\)
(ii) Selling price ₹50
Let the number of dosas sold be \(n\).
Income \(=50n\)
Expenses \(=5000+10n\)
Profit: \(50n-(5000+10n)=2000\)
\(40n-5000=2000\)
\(40n=7000\)
\(n=175\)
(ii) He should aim to sell 175 dosas a day.
10Evaluate the following sequence of fractions: \(\dfrac{1}{3},\ \dfrac{(1+3)}{(5+7)},\ \dfrac{(1+3+5)}{(7+9+11)}\) What do you observe? Can you explain why this happens? [Hint: Recall what you know about the sum of the first n odd numbers.]
\(\dfrac13=\dfrac13\)
\(\dfrac{1+3}{5+7}=\dfrac{4}{12}=\dfrac13\)
\(\dfrac{1+3+5}{7+9+11}=\dfrac{9}{27}=\dfrac13\)
Every fraction in the sequence is equal to \(\dfrac13\).
Why this happens
The sum of the first \(n\) odd numbers is \(n^2\).
The numerator is the sum of the first \(n\) odd numbers, so it is \(n^2\).
The denominator is the sum of the next \(n\) odd numbers.
The first \(2n\) odd numbers add up to \((2n)^2=4n^2\).
So the next \(n\) odd numbers add up to \(4n^2-n^2=3n^2\).
The fraction is \(\dfrac{n^2}{3n^2}=\dfrac13\).
Check with \(n=3\): numerator \(9\), denominator \(4\times9-9=27\) ✓
11Karim and the Genie
Karim was taking a nap under a tree. He had a dream about a magical lamp and a genie. He heard a voice saying, “I have come to serve you, Oh master”. He woke up and to his surprise, it was a genie! “Do you want to make money?”, asked the genie. Karim nodded dumbly in bewilderment. The genie continued, “Do you see the banyan tree over there? All you have to do is go around it once. The money in your pocket will double”. “One moment!”, said the genie. “Since I am bringing you great riches, you should share some of your gains with me. You must give me 8 coins each time you go around the tree.” Thinking that was a trifling amount, Karim readily agreed. He went around the tree once. Just as the genie had said, the number of coins in his pocket doubled! He gave 8 coins to the genie. He made another round. Again the number of coins doubled. He gave 8 more coins to the genie. He went around the tree for the third time. The number of coins doubled again, but to his horror, he was left with only 8 coins, exactly the number of coins he owed the genie! As Karim began to wonder how the genie tricked him, the genie let out a loud laugh and disappeared.
(i) How many coins did Karim initially have?
(ii) For what cost per round should Karim agree to the deal, if he wants to increase the number of coins he has?
(iii) Through its magical powers, the genie knows the number of coins that Karim has. How should the genie set the cost per round so that it gets all of Karim’s coins?Karim goes around the banyan tree; the genie sits on top
Karim was taking a nap under a tree. He had a dream about a magical lamp and a genie. He heard a voice saying, “I have come to serve you, Oh master”. He woke up and to his surprise, it was a genie! “Do you want to make money?”, asked the genie. Karim nodded dumbly in bewilderment. The genie continued, “Do you see the banyan tree over there? All you have to do is go around it once. The money in your pocket will double”. “One moment!”, said the genie. “Since I am bringing you great riches, you should share some of your gains with me. You must give me 8 coins each time you go around the tree.” Thinking that was a trifling amount, Karim readily agreed. He went around the tree once. Just as the genie had said, the number of coins in his pocket doubled! He gave 8 coins to the genie. He made another round. Again the number of coins doubled. He gave 8 more coins to the genie. He went around the tree for the third time. The number of coins doubled again, but to his horror, he was left with only 8 coins, exactly the number of coins he owed the genie! As Karim began to wonder how the genie tricked him, the genie let out a loud laugh and disappeared.
(i) How many coins did Karim initially have?
(ii) For what cost per round should Karim agree to the deal, if he wants to increase the number of coins he has?
(iii) Through its magical powers, the genie knows the number of coins that Karim has. How should the genie set the cost per round so that it gets all of Karim’s coins?
(i) Initial coins
Let Karim have \(x\) coins at the start.
After round 1: \(2x-8\).
After round 2: \(2(2x-8)-8=4x-24\).
In round 3 the coins double to \(2(4x-24)=8x-48\).
This is exactly 8, the coins he owes.
\(8x-48=8\)
\(8x=56\)
\(x=7\)
Check with \(x=7\):
| Round | Coins at start | After doubling | After paying 8 |
|---|---|---|---|
| 1 | 7 | 14 | 6 |
| 2 | 6 | 12 | 4 |
| 3 | 4 | 8 | 0 |
In round 3 he has exactly 8 after doubling, all of which he owes, so he is left with 0 ✓
(ii) Cost for which Karim gains
Let the cost per round be \(c\).
After one round Karim has \(2x-c\).
He gains when \(2x-c\gt x\), that is \(c\lt x\).
If \(c\lt x\), the new amount \(2x-c\) is larger than \(x\), so \(c\) is still less than his coins in the next round.
So his coins keep increasing every round.
In the story \(c=8\) is more than his 7 coins, so he lost.
(iii) Cost for which the genie gets everything
For 3 rounds with cost \(c\): \(x\to2x-c\to4x-3c\to8x-7c\).
The genie gets all the coins when \(8x-7c=0\).
\(c=\dfrac{8x}{7}\)
For \(x=7\), this gives \(c=8\), exactly as in the story.
(For a single round, the genie would set \(c=2x\).)
(ii) The cost per round must be less than the number of coins Karim starts with.
(iii) For 3 rounds, the cost per round should be \(\dfrac{8x}{7}\), where \(x\) is Karim’s number of coins.
Continue with Ganita Prakash, Part II
Move on to Chapter 7: Area, or head back to the Ganita Prakash hub for the complete chapter list.