Algebra Play | IT

Question 8

Fill the following pyramids:

Question diagram 1
Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question
  • In a number pyramid, the value of each block is equal to the sum of the two adjacent blocks directly below it.
  • If three blocks are arranged with block XX resting on blocks YY and ZZ, then X=Y+ZX = Y + Z.
  • By setting up simple linear equations row by row from top to bottom or bottom to top, we can solve for all missing values.

(i) Fill pyramid (i).

Step 1 · Find the Missing Numbers for Pyramid (i)

Let the missing blocks be labeled as follows:

  • Second row from top: aa (left), 2222 (right)
  • Third row from top: bb (left), cc (middle), dd (right)
  • Bottom row: 44, ee, 66, ff, third row [b, c, d], and bottom row [4, e, 6, f].] Diagram 1

From the top block

a+22=50  ⟹  a=50−22=28a + 22 = 50 \implies a = 50 - 22 = 28

For the third row blocks

28=b+c28 = b + c c+d=22c + d = 22

Express bb and cc in terms of the bottom row variable ee

b=4+ec=e+6\begin{aligned} b &= 4 + e \\ c &= e + 6 \end{aligned}

Substitute into 28=b+c28 = b + c

28=(4+e)+(e+6)28=10+2e2e=18e=9\begin{aligned} 28 &= (4 + e) + (e + 6) \\ 28 &= 10 + 2e \\ 2e &= 18 \\ e &= 9 \end{aligned}

Now calculate bb, cc, dd, and ff

b=4+9=13b = 4 + 9 = 13 c=9+6=15c = 9 + 6 = 15 15+d=22  ⟹  d=22−15=715 + d = 22 \implies d = 22 - 15 = 7 d=6+f  ⟹  7=6+f  ⟹  f=1d = 6 + f \implies 7 = 6 + f \implies f = 1
Answer

(i) a=28, b=13, c=15, d=7, e=9, f=1a = 28,\, b = 13,\, c = 15,\, d = 7,\, e = 9,\, f = 1

(ii) Fill pyramid (ii).

Step 1 · Find the Missing Numbers for Pyramid (ii)

Let the missing blocks be labeled as follows:

  • Top block: aa
  • Second row: 4040 (left), bb (right)
  • Third row: cc (left), dd (middle), 99 (right)
  • Bottom row: 55, ee, 77, ff, third row [c, d, 9], and bottom row [5, e, 7, f].] Diagram 2

From the bottom right

7+f=9  ⟹  f=9−7=27 + f = 9 \implies f = 9 - 7 = 2

For the block 4040

c+d=40c + d = 40

Express cc and dd in terms of ee

c=5+ed=e+7\begin{aligned} c &= 5 + e \\ d &= e + 7 \end{aligned}

Substitute into c+d=40c + d = 40

(5+e)+(e+7)=4012+2e=402e=28e=14\begin{aligned} (5 + e) + (e + 7) &= 40 \\ 12 + 2e &= 40 \\ 2e &= 28 \\ e &= 14 \end{aligned}

Now calculate cc, dd, bb, and aa

c=5+14=19c = 5 + 14 = 19 d=14+7=21d = 14 + 7 = 21 b=d+9=21+9=30b = d + 9 = 21 + 9 = 30 a=40+b=40+30=70a = 40 + b = 40 + 30 = 70
Answer

(ii) a=70, b=30, c=19, d=21, e=14, f=2a = 70,\, b = 30,\, c = 19,\, d = 21,\, e = 14,\, f = 2

(iii) Fill pyramid (iii).

Step 1 · Find the Missing Numbers for Pyramid (iii)

Let the missing blocks be labeled as follows:

  • Top block: 3535
  • Second row: aa (left), bb (right)
  • Third row: cc (left), dd (middle), 77 (right)
  • Bottom row: 33, 55, ee, ff, third row [c, d, 7], and bottom row [3, 5, e, f].] Diagram 3

From the bottom left

c=3+5=8c = 3 + 5 = 8

For the top block 3535

a+b=35a + b = 35

Express aa and bb in terms of cc and dd

a=c+db=d+7\begin{aligned} a &= c + d \\ b &= d + 7 \end{aligned}

Substitute into a+b=35a + b = 35

(c+d)+(d+7)=35c+2d+7=35\begin{aligned} (c + d) + (d + 7) &= 35 \\ c + 2d + 7 &= 35 \end{aligned}

Since c=8c = 8

8+2d+7=3515+2d=352d=20d=10\begin{aligned} 8 + 2d + 7 &= 35 \\ 15 + 2d &= 35 \\ 2d &= 20 \\ d &= 10 \end{aligned}

Now calculate ee, ff, aa, and bb

5+e=d  ⟹  5+e=10  ⟹  e=55 + e = d \implies 5 + e = 10 \implies e = 5 e+f=7  ⟹  5+f=7  ⟹  f=2e + f = 7 \implies 5 + f = 7 \implies f = 2 a=c+d=8+10=18a = c + d = 8 + 10 = 18 b=d+7=10+7=17b = d + 7 = 10 + 7 = 17
Answer

(iii) a=18, b=17, c=8, d=10, e=5, f=2a = 18,\, b = 17,\, c = 8,\, d = 10,\, e = 5,\, f = 2

Common Mistakes
  • Double-counting Middle Blocks: Forgetting that an inner block in any row contributes to both blocks directly above it on the left and right.
  • Sign Errors in Linear Equations: Adding instead of subtracting when isolating variables (e.g., writing 2e=40+122e = 40 + 12 instead of 2e=40−122e = 40 - 12).

More questions in IT

Q1

Context: Think of a number trick:

  1. Think of a number.
  2. Double it.
  3. Add four.
  4. Divide by two.
  5. Subtract the original number you thought of.

Q. I predict you get 2. Am I right? Try it out with different starting numbers. Do you always end up with the same value, 2? Why?

Q2

Context: Consider the following number trick:

  1. Think of a number.
  2. Double it.
  3. Add four.
  4. Divide by two.
  5. Subtract the original number you thought of. (This trick always results in 2).

Q. How would you change this game to make the final answer 3? What about 5?

Q3

Context: Consider the following number trick:

  1. Think of a number.
  2. Double it.
  3. Add four.
  4. Divide by two.
  5. Subtract the original number you thought of. (This trick always results in 22).

Q. Can you come up with more complicated steps that always lead to the same final value?

Q4

Context: In the date trick, the final answer is given by 100M+165+D100M + 165 + D, where MM is the month and DD is the day.

Q. Find the dates if the final answers are the following:

(i) 1269

(ii) 394

(iii) 296

Q5

Context: In the date trick, the final answer is given by 100M+165+D100M + 165 + D, where MM is the month and DD is the day.

Q. 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.

Q6

Try to devise your own 'Think of a Number' trick.

Q7

Use the same rule to fill these pyramids:

Q8

Fill the following pyramids:

Q9

What is the relationship between the numbers in the bottom row and the number at the top?

Let us start with the simplest pyramid.

Q10

What about a pyramid with three rows?

Using letter numbers for the bottom row, we can write an expression for the top row.

Q11

In the following grids, find the values of the shapes and fill in the empty squares:

Q12

Context: 6.5 The Largest Product

Q. Fill the digits 2, 3, and 5 in □□×□\square\square \times \square, using each digit once. What is the largest product possible?

Q13

Context: Mukta's trick: Choose a 2-digit number of different digits, reverse the digits to get another number, find their difference, and divide the result by 9.

Q. If we choose other 2-digit numbers, and follow the steps, will there always be no remainder?

Q14

Context: Suppose a two-digit number is abab. When it is reversed, the new number is baba. If b>ab > a, the difference is ba−ab=9(b−a)ba - ab = 9(b - a), which is divisible by 9.

Q. Can you work out what happens if a>ba > b?

← Back to Algebra Play