Question

13. The sum of the squares of the digits constituting a two-digit number is
10, and the product of the required number by the number consisting of
the same digits written in the reverse order is 403. Find the number.

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Answer 1

Answer:

Explanation:

Let n(a,b) be two digit number

Sum of squares of digits =10

a2+b2=10

∴n(a,b)=10a+b

a>0,b>0

(number)×(number with same digits in reverse order)=403

n(a,b)×n(b,a)=403

(10a+b)(10b+a)=403

10(a2+b2)+(101ab)=403

100+101ab=403(∵a2+b2=10)

ab=3

(a+b)2=a2+b2+2ab=16

a+b=4(∵a>0,b>0)

(a−b)2=a2+b2−2ab=4

a−b=±2

a=3,b=1  (or)  a=1,b=3

∴  13  (or)  31  is the required number

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