what is the smallest 5 digit number which when divided by 7, 11 and 21 leave a remainder of 3 in each case
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Answer:
10167
Step-by-step explanation:
LCM of 7,11 and 21 = 231
smallest 5 digit number = 10000
multiple of 231 near 10000 = 10164
so, the required number will be
= 10164 +3
= 10167
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