Question

In a four digit number, the sum of the digits in the thousands and tens is equal to 4, the sum of the digits in the hundreds and the units is 15, and the digit of the units exceeds by 7 the digit of the thousands, among all the numbers satisfying these conditions, find the number the sum of the product of whose digit of the thousands by the digit of the units and the product of the digit of the hundreds by that of the tense assumes the least value.

Answer 1

Answer:

let the number be abcd

⟹a+c=4,b+d=15,d−a=7

a can be 1,2,3,4 from a+c=4 but a cannot be 3 and 4 from d-a=7 as d exceeds 9 which is imposible

for a=1 the number is 1738 and for a=2 the number is 2629 ∵the required number is 1738

Answer 2

Step-by-step explanation:

let the number be abcd

⟹a+c=4,b+d=15,d−a=7

a can be 1,2,3,4 from a+c=4 but a cannot be 3 and 4 from d-a=7 as d exceeds 9 which is imposible

for a=1 the number is 1738 and for a=2 the number is 2629  ∵the required number is 1738