## Problem Of The Week 26

**A three digit number is such that the sum of the digit in the hundred’s place and the ten’s place is 1 more than the digit in the unit’s place. It is also given that the digit in the ten’s place exceeds the square of the digit in the hundred’s place by 1, and that the square of the digit in the units place diminished by 7 is the same as the sum of the squares of the other two digits. What is the number?**

**(1) 346 (2) 256 (3) 458 (4) 526 (5) None of These.**

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option 2 256

Worked through the options 😛

Sorry:)

milindSeptember 25, 2008 at 5:03 pm

the problem is pretty easy let the number be xyz, x can’t be zero

x+y=z+1

y=x^2+1

z^2-7=x^2+y^2

subtract first two

we get z=x(x+1) as x and z are digits, and x not 0

then obviously z=2,6 as these two are the only solutions possible

but with z=2 z^2-7 is negative , hence rejected

so x=2 z=6 and y=5

our number is 256

outtimedSeptember 25, 2008 at 10:20 pm

the problem is pretty easy let the number be xyz, x can’t be zero

x+y=z+1

y=x^2+1

z^2-7=x^2+y^2

subtract first two

we get z=x(x+1) as x and z are digits, and x not 0

then obviously z=2,6 as these two are the only solutions possible

but with z=2 z^2-7 is negative , hence rejected

so x=2 z=6 and y=5

our number is 256

outtimedSeptember 25, 2008 at 10:21 pm