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Arithmetic Smart Math

[Smart Math] Arithmetic Problem 7

Here’s and example of a SMART MATH problem for ARITHMETIC.

Arithmetic

Problem

A number when divided by 2, 3, 5, 6, 7, 11, 13 and 14 leaves the reminders 1, 2, 4, 5, 6, 10, 12 and 13 respectively. What is the number?

  1. 30148
  2. 30029
  3. 31019
  4. 30025
  5. 30229

The Usual Method

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First find the LCM of 2, 3, 5, 6, 7, 11, 13 and 14

i.e 2 x 3 x 5 x 7 x 11 x 13 = 30030.

Since all the numbers leave a remainder equal to 1 less than the divisor; subtract 1 from 30030 = 30029

(Ans: 2)

Estimated Time to arrive at the answer = 45 seconds.

Using Technique

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Option ‘1’ and ‘4’ can be eliminated since, the answer cannot be an even number nor can it end in 5, else it will not leave any remainder when divided by 2 or 5 respectively.

Option ‘5’ is also eliminated since when it is divided by 3, it leaves a remainder 1 instead of 2.
(\because 3+0+2+2+9=16 and 16/3 = 5 + 1 as remainder) Finally subtract 6 from the remaining options and check for divisibility of 7. You will find that 30029 – 6 = 30023 is divisible by 7 and hence option ‘2’ is the answer.

(Ans: 2)

Estimated Time to arrive at the answer = 15 seconds.
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