RRB Junior Engineer Stage 1 Model Papers 8

Question : 21 of 100

Marks: +1, -0

The least number, which is to be added to the greatest number of 4 digits so that the sum may be divisible by 345, is

Solution:

The largest 4-digit number = 9999

345)9999(28

\;\;\;\;\;\;\;\;690

\;\;\;\;\;\;\ ───

\;\;\;\;\;\;\;\;3099

\;\;\;\;\;\;\;\;2760

\;\;\;\;\;\;\ ───

\;\;\;\;\;\;\;\;339

∴ Required number

= 345 – 339 = 6

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