Sum of two 2-digit numbers?
2012-04-01
If the two-digit integers `M` and `N` are positive and have the same digits, but in reverse order, which of the following CANNOT be the sum of `N` and `M`?
A) 181
B) 165
C) 121
D) 99
E) 44
[Spoiler Below]
Without loss of generality, le's say `M` has `a` as the tens digit, and `b` as the units digit.
Consequently, `N` would have `b` as the tens, and `a` as the units digit.
remembering how the decimal system works:
`M = 10*a + b`
`N = 10*b + a`
(adding these two equations)
`M + N = 11*a + 11*b = 11*(a + b)`
since `a` and `b` are both integers, `a + b` must also be an integer, so `M + N = 11* an_integer`
so M + N is a multiple of 11.
A) `181` is prime
B) `165 = 11 * 5 * 3`
C) `121 = 11 * 11`
D) `99 = 11 * 3^2`
E) `44 = 11 * 4`
(A) is the only choice that isn't a multiple of 11, and therefore cannot be the the sum of `M` and `N`.
A) 181
B) 165
C) 121
D) 99
E) 44
[Spoiler Below]
Without loss of generality, le's say `M` has `a` as the tens digit, and `b` as the units digit.
Consequently, `N` would have `b` as the tens, and `a` as the units digit.
remembering how the decimal system works:
`M = 10*a + b`
`N = 10*b + a`
(adding these two equations)
`M + N = 11*a + 11*b = 11*(a + b)`
since `a` and `b` are both integers, `a + b` must also be an integer, so `M + N = 11* an_integer`
so M + N is a multiple of 11.
A) `181` is prime
B) `165 = 11 * 5 * 3`
C) `121 = 11 * 11`
D) `99 = 11 * 3^2`
E) `44 = 11 * 4`
(A) is the only choice that isn't a multiple of 11, and therefore cannot be the the sum of `M` and `N`.