Given
p = POSITIVE, ODD and INTEGER
what is the remainder of p/4?
1) p/8 leaves remainder 5
or p=8q+5
substitute q=0,1,2,3 we get 5,13,21 and so on
dividing each of the above with 4 gives us 1 hence SUFF
2) p=a^2 +b^2
given p is odd
we know E=O+E or E+O
substitute integers in the above like
a=2 and b=3
p=4+9=13 divide by4 gives us 1
a=3 and b=4
p=9+16=25 divide by4 gives us 1
Hence SUFF
Hence D
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integer
Source: Beat The GMAT — Data Sufficiency |
For 1) once you get p = 8q + 5, it is even faster to realize that this is equal to
P = 4(2q) + 5, meaning P is a multiple of 4 with remainder 5.
But recall that you can only have remainders of 0, 1, 2, and 3 when dividing something by 4. So turn the equation in to
P = 4(2q) + 4 + 1.
Since 4 is just another multiple of 4, remainder = 1.
P = 4(2q) + 5, meaning P is a multiple of 4 with remainder 5.
But recall that you can only have remainders of 0, 1, 2, and 3 when dividing something by 4. So turn the equation in to
P = 4(2q) + 4 + 1.
Since 4 is just another multiple of 4, remainder = 1.
Also, As for 2),
since Odd = Even +ODD
P = (2k)^2 + (2k+1)^2
4k^2+4k^2+4k+1
First three terms are all multiples of 4, thus leaving 1 as remainder.
since Odd = Even +ODD
P = (2k)^2 + (2k+1)^2
4k^2+4k^2+4k+1
First three terms are all multiples of 4, thus leaving 1 as remainder.
