Question

que please provide method for que 31 and 32

Answer 1

Heya !!!

This is. your answer......

1). The least number that is divisible by all the numbers from 1 to 10.



Solution:-----
To find the least number, we find the LCM of those numbers.

So, LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 is.......

1 = 1
2 = 1×2
3 = 1×3
4 = 1×2×2
5 = 1×5
6 = 1×2×3
7 = 1×7
8 = 1×2×2×2
9 = 1×3×3
10 = 1×2×5

Hence, LCM = 1×2×2×2×3×3×5×7 = 2520.

Hence, the least number which is exactly divisible by all the numbers from 1 to 10 is 2520.



2). The largest number which divides 70 and 125, leaving remainders, 5 and 8 respectively.



Solution:----
Given numbers :-
70 with remainder 5.
So, 70-5 = 65.

125 with remainder 8.
So, 125-8 = 117.

Now, taking HCF of 65 and 117.

65 = 1×5×13
117 = 1×3×3×13

HCF = 1×13

Hence, the largest number which divides 70 and 125, leaving remainder 5 and 8 respectively is 13.

Hope it helps.......

Related :-
#The least number that is divisible by all the numbers from 1 to 10.

#The largest number which divides 70 and 125, leaving remainders, 5 and 8 respectively.