When 860, 1712 and 3203 are divided by a two-digit number $$x$$​, then the remainder in each case is $$y$$​. If $$(2x+y)$$​ is written as $$p×q×r^2$$​, where $$p,q$$​ and $$r$$​ are prime numbers, then what is the value of $$(p+q+r)$$​?

Correct option is A

Given:

• When 860, 1712, and 3203 are divided by a two-digit number x, they all leave the same remainder y
• That means:
• 860 ≡ y (mod x)
• 1712 ≡ y (mod x)
• 3203 ≡ y (mod x)

From this, we know:

• (1712 − 860) = 852 is divisible by x
• (3203 − 1712) = 1491 is divisible by x
• (3203 − 860) = 2343 is divisible by x

So, x must be a common divisor of 852, 1491, and 2343

Step 1: Find HCF of 852, 1491, and 2343

Find HCF(852, 1491):

• 852 = 2 × 2 × 3 × 71
• 1491 = 3 × 3 × 3 × 3 × 3 × 11 (or 3 × 497)

Common factor: 3

Try dividing both:
852 ÷ 3 = 284
1491 ÷ 3 = 497

No higher common factor in 284 and 497, so HCF = 3

Now check HCF(3, 2343)
2343 ÷ 3 = 781

So, HCF = 3

But 3 is not a two-digit number.

Try higher common divisors using Euclidean Algorithm:

Try Euclidean method:

HCF(852, 1491)
1491 ÷ 852 = 1, remainder = 639
852 ÷ 639 = 1, remainder = 213
639 ÷ 213 = 3, remainder = 0
=> HCF = 213

Now check HCF(213, 2343):

2343 ÷ 213 = 11 exactly

So, x = 213

x is a three-digit number, but the question says x is a two-digit number.

Let’s retry carefully — we need common divisors of differences:

Step 2: Find common divisors of 852, 1491, 2343

List of factors of 852, 1491, 2343:

• HCF(852, 1491, 2343) = 213

We try divisors of 213:
213 = 3 × 71 => Try 71

Check:

• 860 ÷ 71 = 12 remainder 8
• 1712 ÷ 71 = 24 remainder 8
• 3203 ÷ 71 = 45 remainder 8

So, x = 71, y = 8

Now we need to find:
(2x + y) = 2×71 + 8 = 150

Then express 150 = p × q × r², where p, q, r are primes.

Factor 150:
150 = 2 × 3 × 5²

So:

• p = 2
• q = 3
• r = 5

Now compute:
p + q + r = 2 + 3 + 5 = 10

Final Answer: (A) 10