Question

5. Find the smallest number which when increased by 7 is exactly divisible by 6 & 32.

Answer 1

89+7=96
96/6=16
96/32=3

Answer 2

Answer:

The smallest number which when increased by 7 is exactly divisible by 6 and 32 is 89.

Step-by-step explanation:

To find: The smallest number which both 32 and 6 can divide. To find such number we have to first find the LCM of the two numbers that are given:

LCM of the number (6, 32) = 96

The number to which if 7 is added 96

Therefore, the number is 96 – 7 = 89

So the smallest number which when increased by 7 to divide both 6 and 32 is 89.