Question

1. If both 112 and 34 are factors of the number a * 43 * 62 * 1711, then what is the smallest possible value of a? A) 121 b) 3267 c) 363 d) 1089

Answer 1

Answer:

952

Step-by-step explanation:

Let x = a*43*62*1711

Writing x in terms of prime factors

43 = 1*43

62 = 2*43

1711 = 29*59

So, x = a*(43)*(2*31)*(59*29)

Given 112 = 2⁴*7 to divide x, each of the factor should be

contained in x

Given 34 = 2*17 to divide x, each of the factor should be

contained in x

Since 16 is a factor in 112, for 112 to divide x , x should also have

a factor of 16. But there is only one 2 in 62, hence there should

be at least 3 more two's in 'a' so that x is divisible by 16.

Since 7 is a factor in 112, for 112 to divide x , x should also have a

factor of 7. But there is no factor of 7, hence there should be at

least one 7 in 'a' so that  x is divisible by 7.

Since 17 is a factor in 34, for 34 to divide x , x should also have a

factor of 17. But there is no factor of 17, hence there should be at

least one 17 in 'a' so that  x is divisible by 17.

Hence, 'a' should contain at least 2³*7*17 = 952 as factor

The smallest possible value of 'a' is 952.

Hope, it helps !