Class 6 Playing with Numbers Exercise 2.10-1

\begin{array}{l} \text {Q1. What is the smallest number which when divided by 24,36 and 54 } \\ \text {gives a remainder of 5 each time? } \\ \\\text {Sol. Prime factorization of 24 } =2 \times 2 \times 2 \times 3 \\ \text {Prime factorization of 36 }=2 \times 2 \times 3 \times 3 \\ \text {Prime factorization of 54 }=2 \times 3 \times 3 \times 3 \\\therefore \text {LCM of 24,36 and 54 } \\ =2 \times 2 \times 2 \times 3 \times 3 \times 3 \\ =216 \\\text {The smallest number which is exactly divisible by 24,36 and 54 is 216.} \\\text {To get remainder as 5, we need to add 5 to 216 } =216+5=221 \\ \therefore \text {The smallest number which when divided by 24,26 and 54 gives } \\ \text { a remainder of 5 each time is 221.} \\ \\\text {Q2. What is the smallest number that both 33 and 39 divide leaving remainders of 5 ? } \\ \\\text {Sol. Prime factorization of 33 } =3 \times 11 \\ \text {Prime factorization of 39 }=3 \times 13 \\\therefore \text {LCM of 33 and 39 } \\ =3 \times 11 \times 13 =429 \\\text {The smallest number which is exactly divisible by 33 and 39 is 429.} \\\text {To get remainder as 5, we need to add 5 to 429 } =429+5=434 \\ \therefore \text {The smallest number which when divided by 33 and 39 gives } \\ \text { a remainder of 5 each time is 434.} \\ \\\text {Q3. Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive) } \\ \\ \end{array}\begin{array}{l} \text {Sol. We need to find the LCM of numbers from 1 to 10.} \\ \text {We know that 2,3,5 and 7 are prime numbers.} \\\text {Prime factorization of 4}=2 \times 2\\ \text {Prime factorization of 6}=2 \times 3\\ \text {Prime factorization of 8}=2 \times 2 \times 2\\ \text {Prime factorization of 9}=3 \times 3\\ \text {Prime factorization of 10}=2 \times 5\\ \therefore \text {LCM of numbers from 1 to 10 } \\ =2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 \\ =2520 \\ \therefore \text {2520 is the least number that is divisible by all the numbers between 1 and 10.} \\ \\\text {Q4. What is the smallest number that, when divided by 35,56 and 91 } \\ \text {leaves remainders of 7 in each case? } \\ \\\text {Sol. Prime factorization of 35 } =5 \times 7 \\ \text {Prime factorization of 56 }=2 \times 2 \times 2 \times 7 \\ \text {Prime factorization of 91 }=7 \times 13 \\\therefore \text {LCM of 35,56 and 91 } \\ =2 \times 2 \times 2 \times 5 \times 7 \times 13 \\ =3640 \\\text {The smallest number which is exactly divisible by 35,56 and 91 is 3640.} \\\text {To get remainder as 7, we need to add 7 to 3640 } =3640+7=3647 \\ \therefore \text {The smallest number which when divided by 35,56 and 91 gives } \\ \text { a remainder of 7 each time is 3647.} \\ \\\text {Q5. In a school there are two sections- section A and section B of Class VI.} \\ \text {There are 32 students in section A and 36 in section B. Determine the minimum } \\ \text { number of books required for their class library so that they can be distributed } \\ \text {equally among students of section A or section B.} \\ \\\end{array}\begin{array}{l} \text {Sol. } \\\text {Prime factorization of 32}=2 \times 2 \times 2 \times 2 \times 2 \\ \text {Prime factorization of 36}=2 \times 2 \times 3 \times 3 \\\therefore \text {LCM of 32 and 36 } \\ =2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \\ =288 \\\text {Hence, the minimum number of books required for their class library } \\ =\text {LCM of 32 and 36} =288 \text { books.} \\ \\ \end{array}