Question 550112: The natural number n is the smallest number satisfying the following properties: when divided by 3 remainder is 1,when divided by 4 remainder is 2,when divided by 5 remainder is 3,when divided by 6 remainder is 4,what is the remainder in case of division by 7?
Found 3 solutions by ankor@dixie-net.com, fcabanski, richard1234:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! The natural number n is the smallest number satisfying the following properties: when divided by 3 remainder is 1,
when divided by 4 remainder is 2,
when divided by 5 remainder is 3,
when divided by 6 remainder is 4,
what is the remainder in case of division by 7?
:
I can't think of an algebra equation that can solve this but we can use some logic here.
Since one of the requirements is divide by 5, remainder 3, we know the number has a unit value of 3 or 8
Also if it meets the requirement of divide by 6, remainder 4, it will automatically satisfy the requirement of divide 3, remainder 1
This narrows it down, I wrote numbers;
18, 23, 28, 33, 38, 43, 48, 53, 58, 63
easy to throw out 18, 28, 33, 48, 63
tried, 23, 38, 43, 53, then came the winner 58!!
:
"what is the remainder in case of division by 7"
58/7 = 8 remainder 2
:
I am thinking there may be a more efficient way to do this, but just can't come up with it this Christmas Eve. Merry Christmas!!
Answer by fcabanski(1391)  (Show Source):
You can put this solution on YOUR website! Remainders for different numbers divided by different numbers run in patterns. Start with 3.
4/3 = 1 R 1, 5/3= 1 R2, 6/3 = 2, 7/3=2R1, 8/3 = 2R2. Starting from 4 every 3rd number has a remainder of 1 when divided by 3: 4, 7, 10, 13....
Find the pattern of remainders when dividing those by 4.
4/4 = R0, 7/4 = 1R3, 10/4=R2, 13/4=R1, 16/4=R0...22/4=R2...every 12 in the 3's remainder 1 series will have a remainder of 2 when divided by 4. So that's 10,22,34,46,58...
Find the pattern of remainders when dividing those by 5.
10/5=R0, 22/5=R2, 34/5=R4, 46/5=R1, 58/5=R3, 70/5=R0, 82/5=R2, 94/5=R4, 106/5=R1, 118/5=R3...every 60 in the 4's series will have remainder 3 when divided by 5. So that's 58, 118, 178, 238, 298...
Find the pattern of remainders when dividing those by 6.
58/6=9R4. That's the smallest number that has remainder 1 when divided by 3, remainder 2 when divided by 4, R3 when divided by 5, and R4 when divided by 6.
58/7=9 R 2
If you need help understanding math so you can solve these problems yourself, then one on one online tutoring is the answer ($30/hr). If you need faster solutions with guaranteed detailed answers, then go with personal problem solving ($3.50-$5.50 per problem). Contact me at fcabanski@hotmail.com
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! Here I use "mod" to shorten the solution. "Mod" or "modulo" is another way of expressing remainders.
The number is 3 mod 5 so the units digit must be either 3 or 8. We also know the number is 2 mod 4, so it must be even but not divisible by 4. Hence the units digit is 8. Also, it is 1 mod 3 so 28, 58, 88, 118, ... are the choices. 58 is the smallest positive integer that works, and 58 is congruent to 2 mod 7, so the remainder is 2.
|
|
|
