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What is the greatest number of six digits, which when divided by each of 16, 24, 72 and 84, leaves the remainder 15?
999981
999951
999963
999915
- To solve this, we seek the largest 6-digit number which leaves a remainder 15 when divided by 16, 24, 72, and 84.
- Let the number be N. Then, N = K × LCM(16, 24, 72, 84) + 15, where N is a 6-digit number.
- First, the least common multiple (LCM) of 16, 24, 72, and 84 has to be found:
- 16 = 24
- 24 = 2³ × 3
- 72 = 2³ × 3²
- 84 = 2² × 3 × 7
- LCM = 24 × 3² × 7 = 16 × 9 × 7 = 1008
- So, N = (multiple of 1008) + 15 produces the required remainders.
- The largest 6-digit number is 999999.
- First, subtract 15: 999999 - 15 = 999984
- Now, 999984 ÷ 1008 = 992.042, so nearest integer multiple is 992
- 1008 × 992 = 999936
- Add 15: 999936 + 15 = 999951
- Let's review the options:
1. 999981 (not in the form 1008n+15)
2. 999951 (matches calculation!)
3. 999963 (not matching 1008n+15)
4. 999915 (does not fit format)
- Correct Answer: Option 2 - 999951
By: santosh ProfileResourcesReport error
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