If ab is a two-digit number and a, b, x,y are digits
satisfying (ab)2 = x! + y, then which of the following
may be true?
Answers
Given: (ab)² = a!+b ab is a two-digit number
To find : a & b
Solution:
a! + b = (ab)²
ab = 10a + b
if a = 0 then 0! + b = b² => b² - b - 1 = 0 ( no real value of b )
=> if a = 1 then 10a + b ≥ 10 => (ab)² ≥ 100 but 1 ! = 1
a = 2 then 10a + b ≥ 20 => (ab)² ≥ 400 but 2 ! = 2
a = 3 then 10a + b ≥ 30 => (ab)² ≥ 900 but 3 ! = 6
a = 4 then 10a + b ≥ 40 => (ab)² ≥ 1600 but 4 ! = 24
a = 5 then 10a + b ≥ 50 => (ab)² ≥ 2500 but 5 ! = 120
a = 6 then 10a + b ≥ 60 => (ab)² ≥ 3600 but 5 ! =720
a = 7 then 80 > 10a + b ≥ 70 => 6400 > (ab)² ≥ 4900 , 7 ! =5040
7! lies between 4900 & 6400
lets take b = 1
then (ab)² = 71² = 5041
a! + b = 5040 + 1 = 5041
5041 = 5041
=> a! + b = (ab)²
Hence a = 7 & b = 1
a = 8 then 90 ≥ 10a + b ≥ 8 0 =>8100 > (ab)² ≥ 6400 but 8 ! =40320
Hence not possible
a = 7 & b = 1 is only solution
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