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# Help with math 2?

When an integer is divided by 5 the remainder is 3. What is the remainder when 4 times that integer is divided by 5?

ans is 2
6 replies

## Replies to: Help with math 2?

• 1775 replies14 threads Senior Member
edited November 2018
5x+3 is the integer, the new equation will be: (5x+3)*4/5
Remainder 3*4=12, 12/5 = 2, with 2 as remainder. so the answer is 2
edited November 2018
• 7722 replies24 threads Senior Member
I just plugged in a number. Took about 5 seconds to use x = 1. That makes I = 8. Then, multiply 8*4/5 = 32/5 = 6 remainder 2. Not very elegant but it works and you can clearly see the remainder.

The tricky part of this problem, at least for me, would be that using nice and neat algebra can trip you up. In the original equation, I = 5x + 3, the constant and the remainder are the same value: 3. In the new equation, 4/5*I = 4/5*(5x+3), the constant is 12/5, or 2.2, or 2 R2. So the remainder is NOT the same thing as the constant. Also, many will unthinkingly start with something like I/5 = x+3 which will start you off on the wrong foot.

• 154 replies14 threads Junior Member
edited April 2019
Step 1: Plug in a number that works for the integer (8)

Step 2: Since 8 is the integer, multiply it by 4 to get the new integer and you get 32.

Step 3: Divide 32 by 5 and you get 6 remainder 2.

Step 4: Since the remainder is 2, the answer is 2.

An alternative way to do this would be to use algebra (set the integer to 5x+3), but there's no need when you can plug in a number that fits the description of the integer given.
edited April 2019