Let the number N be ′6×104+p+k′ where p is awhole number and k is some natural number. According to the information given: 6×104+p+k=25(6×104+p+k−6×104+p)= 25k ⇒ 24k = 6 x 10‌4+P ⇒ k = 2500 x 10 ‌p So N = 62500 x 10‌p. Sum of the digits of N = 13.