How do i solve ?
51) How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?
A) 15
B) 16
C) 17
D) 18
E) 19
2007-12-23 21:01:00 · 4 answers · asked by sanko 1 in Science & Mathematics ➔ Mathematics
the first no is 1 and the last no is 49.the diff betn two nos is 3
therefore by arithmatic progression,
49=1 + (n-1)3
48=(n-1)3
16=(n-1)
n=17
therefore ans is 17.
2007-12-23 21:36:55 · answer #1 · answered by Dhruv C 3 · 1⤊ 0⤋
the first is one . when you divide one by zero , you have 0 with reminder 1. And all integer such as 1+n*3 lower than 50 fulfill the conditions. with n=16 you find 16*3+1=49
So you have 16+1=17 integers
2007-12-24 05:06:49 · answer #2 · answered by maussy 7 · 2⤊ 0⤋
you could do this as follows:
solve first for any numeric solution at the limits:
n*3+1=0; n=-0.3333
n*3+1=50; n=16.3333
then ask how many integers are there in the range -0.3... -> 16.3.... [0 to 16 = 17 integers]
2007-12-24 05:38:19 · answer #3 · answered by mis42n 4 · 1⤊ 0⤋
I vote for the answer of Dhruv C.
http://www.mathalino.com
2007-12-24 05:55:05 · answer #4 · answered by Toralba 2 · 0⤊ 0⤋