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Number Play - Numbers can Tell us Things

Grade 6CBSE

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

🔑Concepts

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Numbers can be written in a generalized form. For example, a two-digit number abab is represented as 10a+b10a + b.

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A three-digit number abcabc is represented as 100a+10b+c100a + 10b + c.

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Divisibility by 1010: A number is divisible by 1010 if its units digit is 00.

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Divisibility by 55: A number is divisible by 55 if its units digit is either 00 or 55.

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Divisibility by 22: A number is divisible by 22 if its units digit is 0,2,4,6,0, 2, 4, 6, or 88.

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

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Divisibility by 99: A number is divisible by 99 if the sum of its digits is divisible by 99.

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Divisibility by 1111: A number is divisible by 1111 if the difference between the sum of digits at odd places and the sum of digits at even places is either 00 or divisible by 1111.

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In cryptarithmetic puzzles, each letter represents a distinct digit from 00 to 99, and the leading digit of a number cannot be 00.

📐Formulae

N=10a+bN = 10a + b

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

Sum of digits of abc=a+b+c\text{Sum of digits of } abc = a + b + c

Difference for divisibility by 11=∣(a+c+...)−(b+d+...)∣\text{Difference for divisibility by 11} = |(a + c + ...) - (b + d + ...)|

💡Examples

Problem 1:

Find the value of QQ in the following addition: 31Q+1Q3501\begin{array}{r} 31Q \\ + 1Q3 \\ \hline 501 \end{array}

Solution:

Looking at the units column: Q+3=11Q + 3 = 11 (since it must end in 11). Therefore, Q=11−3=8Q = 11 - 3 = 8. Let us check the tens column: 1+Q+11 + Q + 1 (carry) =1+8+1=10= 1 + 8 + 1 = 10. This gives 00 in the tens place and a carry of 11. In the hundreds column: 3+1+1=53 + 1 + 1 = 5. Thus, Q=8Q = 8.

Explanation:

In addition puzzles, we start from the units column and account for carries to the next column.

Problem 2:

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

Solution:

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+y= 2 + 1 + y + 5 = 8 + y. For 8+y8 + y to be divisible by 99, 8+y8 + y must be 99 or 1818 etc. If 8+y=98 + y = 9, then y=1y = 1. If 8+y=188 + y = 18, then y=10y = 10, which is not a single digit. Thus, y=1y = 1.

Explanation:

We apply the divisibility rule for 99 and solve for the unknown digit within the range 00 to 99.

Problem 3:

Check if 13311331 is divisible by 1111.

Solution:

Sum of digits at odd places (from right): 1+3=41 + 3 = 4. Sum of digits at even places (from right): 3+1=43 + 1 = 4. Difference =4−4=0= 4 - 4 = 0. Since the difference is 00, 13311331 is divisible by 1111.

Explanation:

The rule for 1111 involves alternating the sum of digits and checking if the result is a multiple of 1111 or zero.