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Find the least number such that the remainder when divided by 17 is 3, when divided by 16 is 10, and when divided by 15 is 0.
If I have the question correctly stated, then the answer from your book is wrong since 3910/15 is not a whole number.
(1) The number must be a multiple of 15 since the remainder is 0.
(2) We look for the first multiple of 15 that is 10 more than a multiple of 16. We find that 90=15*6 and 16*5+10=90. The LCM of 15 and 16 is 240, so the answer will be a number of the form 90+240x.
(3) We need a number of the form 90+240x that is 3 more than a multiple of 17. We notice that 90=17*5+5, 330=17*19+7 so we try the other direction and see that -150=17*(-9)+3. We presumably want the smallest positive number or the problem is unanswerable, so we add the LCM of 240 and 17 which is 4080 to get 3930.
Note that 3930=15*262, 3930=16*245+10, and 3930=17*231+3 so 3930 is the solution.
If you are studying modular arithmetic, note that the number x has the following properties:
So we are looking for a w such that
15w=10(mod16) or w=6(mod16)
15w=3(mod17) or 5w=1(mod17)
Solving for w yields w=262 and 15(262)=3930, the number we are seeking.
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