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Secondary 1 | Maths
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Date Posted: 4 years ago
Views: 174
No matter the choice, there will always be a remainder of 1.
If there had been 1 apple fewer, there will never be any leftover at all.
Thus, we look for numbers which are common multiples of 2, 3, 4, 5, 6, 7, 8 and 9.
The lowest among them is 2520 (9 x 8 x 7 x 5, since 2 and 4 are contained within 8 while 6 and 3 are contained inside as well)
With the 1 extra apple, he can possibly have 2521 apples.
If there had been 1 apple fewer, there will never be any leftover at all.
Thus, we look for numbers which are common multiples of 2, 3, 4, 5, 6, 7, 8 and 9.
The lowest among them is 2520 (9 x 8 x 7 x 5, since 2 and 4 are contained within 8 while 6 and 3 are contained inside as well)
With the 1 extra apple, he can possibly have 2521 apples.
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