Solution

If HCF of two numbers is 26, then the two numbers are 26X & 26Y where X & Y are co-prime numbers. ⇒ H.C.F(X,Y) = 1 Now the sum of the two numbers is 156. So we can say, 26X + 26Y = 156 ⇒ X + Y = 6 The possible pairs of X & Y are: (1, 5), (2, 4), (3, 3), (5, 1), (4, 2) [As we are not considering the ordered pairs (2, 4), (4, 2), (3, 3) because X and Y are coprime] Accepted possible pairs of X & Y are: (1, 5), (5, 1) Such a number of pairs, For  (1, 5) => 26,  (26 × 5) = 130 For  (5, 1) => (26 × 5) = 130, 26 Combining the above results, So, there will be only one pair of numbers which will give the sum of two numbers is 156 and their HCF is 16 and that pair of the numbers will be (26, 130) only.