A Four Digit Number

Number Theory Level 3

Find a four-digit number which leaves a remainder of 21 when divided by 133, and
a remainder of 6 when divided by 134.


The answer is 2016.

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Paul Fournier by Paul Fournier , 70, Canada

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2 solutions

Paul Fournier
Jul 29, 2016

Relevant wiki: Modular Arithmetic - Word Problems

  1. a=133 x q + 21
  2. a=134 x t + 6
  3. 133q+21=134t+6
  4. 15=133t-133q+t
  5. 15=133(t-q)+t
  6. The largest and the smallest 4 digit numbers are 9999 and 1000 thus 9999/134=74.6... 1000/134=7.46...
  7. t and q are between 8 and 74 .
  8. from 5 the only possibility is if t=q=15
  9. 133x15+21=2016
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DarkMind S.
Aug 17, 2016

N / 133 = Q + 21

N / 134 = Q + 6

N = 133Q + 21

N = 134Q + 6

Now, 133Q + 21 = 134Q + 6 Q = 15

Now, . N = 133( 15 ) + 21

N = 134(15 ) + 6

N =. 2016

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