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Help with math 2?

student3737student3737 Registered User Posts: 8 New Member
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

Replies to: Help with math 2?

  • makemesmartmakemesmart Registered User Posts: 1,186 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
  • JBStillFlyingJBStillFlying Registered User Posts: 6,355 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.

  • skompella9892skompella9892 Registered User Posts: 150 Junior Member
    edited April 30
    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.
  • RichInPittRichInPitt Registered User Posts: 464 Member
    If you have 3 left after dividing a number by 5, then you have 4*3 = 12 left after dividing four of that number by 5. Then that 12 can be divided by 5 with 2 left over.
  • bjkmombjkmom Registered User Posts: 7,976 Senior Member
    This question was posted in November. I'm guessing he has his answer, or no longer cares.
  • RichInPittRichInPitt Registered User Posts: 464 Member
    Math test occur every year, mathematical approaches to solutions don't change, people read these forums for years and might get value from reading them. People still learn from Brahmagupta's solutions from 1000 years ago.
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