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IM LOST ON HOW YOU GOT FROM n = (50+10h)/11 TO 4 +h + [(6-h)/11].palvarez wrote:n = (50+10h)/11 = 4 +h + [(6-h)/11]
h = 6 (mod 11)
h = 6 or 17, 28, 39 etc
h = 39, then n = 43-3 = 40
hpgmat wrote:IM LOST ON HOW YOU GOT FROM n = (50+10h)/11 TO 4 +h + [(6-h)/11].palvarez wrote:n = (50+10h)/11 = 4 +h + [(6-h)/11]
h = 6 (mod 11)
h = 6 or 17, 28, 39 etc
h = 39, then n = 43-3 = 40
PLEASE EXPLAIN
palvarez wrote:hpgmat wrote:IM LOST ON HOW YOU GOT FROM n = (50+10h)/11 TO 4 +h + [(6-h)/11].palvarez wrote:n = (50+10h)/11 = 4 +h + [(6-h)/11]
h = 6 (mod 11)
h = 6 or 17, 28, 39 etc
h = 39, then n = 43-3 = 40
PLEASE EXPLAIN
50+10h = 11(4+h) + (6-h)
I GET THAT NOW , THANKS . BUT , HOW DO YOU GET THOSE NUMBERS FOR H?
hpgmat wrote:Here is the logic:palvarez wrote:hpgmat wrote:IM LOST ON HOW YOU GOT FROM n = (50+10h)/11 TO 4 +h + [(6-h)/11].palvarez wrote:n = (50+10h)/11 = 4 +h + [(6-h)/11]
h = 6 (mod 11)
h = 6 or 17, 28, 39 etc
h = 39, then n = 43-3 = 40
PLEASE EXPLAIN
50+10h = 11(4+h) + (6-h)
I GET THAT NOW , THANKS . BUT , HOW DO YOU GET THOSE NUMBERS FOR H?
n = (50+10h)/11 = 4+h + ((6-h))/11)
if n is an integer, (6-h)/11 has to be an integer.
or 6-h is a multiple of 11
or h = 11k+6, for any integer k.
Based on the answer choices pick some number k and match it
