The largest 6-digit number which is exactly divisible by 29 is (a) 9,99,972 (b9,99,974 (c9,99,976 (d9,99,978
Answers
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HERE IS YOUR ANSWER..........
THE TRICK USED HERE IS THAT THE LAST DIGITS MUST BE DIVISIBLE BY 29 . SO IN THESE CASES ,
(D.) OPTION IS RIGHT THE NO. DIVISIBLE BY 29 IS 9,99,978...
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9,99,978 is exactly divisible by 29.
Correct option: d) 9,99,978
Tips:
To cheek the divisibility by 29, we add three times the last digit with the leading number (except the last digit) and check whether divisible by 29.
The result may not come in the first step. The step may have to be repeated.
Step-by-step explanation:
(a) 999972
Here, 999972
➤ 99997 + 3 × 2 = 99997 + 6 = 100003
➤ 10000 + 3 × 3 = 10000 + 9 = 10009
➤ 1000 + 3 × 9 = 1000 + 27 = 1027
➤ 102 + 3 × 7 = 102 + 21 = 123
➤ 12 + 3 × 3 = 12 + 9 = 31, not divisible by 29.
This number is not divisible by 29.
(b) 999974
Here, 999974
➤ 99997 + 3 × 4 = 99997 + 12 = 100009
➤ 10000 + 3 × 9 = 10000 + 27 = 10027
➤ 1002 + 3 × 7 = 1002 + 21 = 1023
➤ 102 + 3 × 3 = 102 + 9 = 111
➤ 11 + 3 × 1 = 11 + 3 = 14, not divisible by 29.
This number is not divisible by 29.
(c) 999976
➤ 99997 + 3 × 6 = 99997 + 18 = 100015
➤ 10001 + 3 × 5 = 10001 + 15 = 10016
➤ 1001 + 3 × 6 = 1001 + 18 = 1019
➤ 101 + 3 × 9 = 101 + 27 = 128
➤ 12 + 3 × 8 = 12 + 24 = 36, not divisible by 29.
This number is not divisible by 29.
(d) 999978
➤ 99997 + 3 × 8 = 99997 + 24 = 100021
➤ 10002 + 3 × 1 = 10002 + 3 = 10005
➤ 1000 + 3 × 5 = 1000 + 15 = 1015
➤ 101 + 3 × 5 = 101 + 15 = 116
➤ 11 + 3 × 6 = 11 + 18 = 29, divisible by 29.
This number is divisible by 29.
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