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Number Play - Is This a Multiple Of?

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A two-digit number with digits aa and bb is represented in generalized form as 10a+b10a + b.

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A three-digit number with digits a,b,a, b, and cc is represented as 100a+10b+c100a + 10b + c.

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Divisibility by 1010: A number is a multiple of 1010 if its ones digit is 00.

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Divisibility by 55: A number is a multiple of 55 if its ones digit is either 00 or 55.

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Divisibility by 22: A number is a multiple of 22 if its ones digit is an even number, i.e., 0,2,4,6,0, 2, 4, 6, or 88.

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Divisibility by 99: A number is a multiple of 99 if the sum of its digits is divisible by 99. If N=abcN = abc, then (a+b+c)(a + b + c) must be a multiple of 99.

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Divisibility by 33: A number is a multiple of 33 if the sum of its digits is divisible by 33.

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Letters for Digits: In puzzles, each letter stands for a single digit (0−9)(0-9), and the first digit of a number cannot be 00.

📐Formulae

N=10a+bN = 10a + b

N=100a+10b+cN = 100a + 10b + c

Sum of digits=d1+d2+d3+⋯+dn\text{Sum of digits} = d_1 + d_2 + d_3 + \dots + d_n

If 9∣(a+b+c)  ⟹  9∣abc\text{If } 9 \mid (a+b+c) \implies 9 \mid abc

If 3∣(a+b+c)  ⟹  3∣abc\text{If } 3 \mid (a+b+c) \implies 3 \mid abc

💡Examples

Problem 1:

If 21y521y5 is a multiple of 99, where yy is a digit, what is the value of yy?

Solution:

y=1y = 1

Explanation:

For a number to be a multiple of 99, the sum of its digits must be divisible by 99. Sum of digits = 2+1+y+5=8+y2 + 1 + y + 5 = 8 + y. For 8+y8 + y to be a multiple of 99, 8+y8 + y must be 9,18,…9, 18, \dots. Since yy is a single digit, 8+y=9  ⟹  y=18 + y = 9 \implies y = 1.

Problem 2:

Check if the number 357357 is divisible by 33.

Solution:

Yes, 357357 is divisible by 33.

Explanation:

Find the sum of the digits: 3+5+7=153 + 5 + 7 = 15. Since 1515 is divisible by 33 (15÷3=515 \div 3 = 5), the number 357357 is also divisible by 33.

Problem 3:

Find the values of the letters AA and BB in the following addition: 3A+25B2\begin{array}{r} 3A \\ + 25 \\ \hline B2 \end{array}

Solution:

A=7,B=6A = 7, B = 6

Explanation:

In the ones column, A+5A + 5 gives a number ending in 22. This means A+5=12A + 5 = 12, so A=7A = 7. Carrying over 11 to the tens column, we get 1+3+2=B1 + 3 + 2 = B, which means B=6B = 6.

Problem 4:

If 31z531z5 is a multiple of 33, where zz is a digit, find all possible values of zz.

Solution:

z=0,3,6,9z = 0, 3, 6, 9

Explanation:

The sum of digits is 3+1+z+5=9+z3 + 1 + z + 5 = 9 + z. For the number to be divisible by 33, 9+z9 + z must be a multiple of 33. Possible values for 9+z9 + z are 9,12,15,189, 12, 15, 18. Thus, zz can be 0,3,6,0, 3, 6, or 99.