What is the sum of the digits of a two digits number , which is less than 32 less than the square of the product of its digits
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Answered by
10
Answer:
Let n(a,b) be two digit number.
n(a,b)=10a+b
Given that, n(a,b)+11=a
2
+b
2
10a+b+11=a
2
+b
2
→(1)
n(a,b)=2ab+5
10a+b=2ab+5→(2)
(1)−(2)⇒11=a
2
+b
2
−2ab−5
(a−b)
2
=16
a−b=±4⇒a=b+4(or)a=b−4
a=b+4,
10b+40+b=2b(b+4)+5
2b
2
−3b−35=0
(2b+7)(b−5)=0
∴ b=5,a=9
a=b−4
10b−40+b=2b
2
−8b+5
2b
2
−19b+45=0
(2b−9)(b−5)=0
∴ b=5,a=1
15 and 95 are the required numbers.
Answered by
9
Answer:
Step-by-step explanation:
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