# Solution

I don't know whether there is an elegant way in which this can be solved; I figured out the solution by using my own clumsy method.

Well, let the number of apples be "x". the first clue tells me that when two apples are transferred, one remains. This obviously implies that the number of apples is an odd number greater than 1; thus, "x" is odd and greater than 1.

Clue number 7 states that when eight apples are transferred at a time, seven apples remain. This implies that x = 8a + 7, where a = 1, 2, 3, etc. So, when a = 1, x = 15. Thus, from this clue we gather that the smallest number of apples in the basket is 15. The basket can't hold more than 600 apples; so, the maximum number of apples possible is 600. Therefore, we need not consider any number larger than 600.

My method, as I have already stated, is a clumsy one. I have laboriously calculated "x" for different values of "a":

Here's what I get: x = 8*1 + 7 = 15; x = 8*2 + 7 = 23; x = 8*3 + 7 = 31; x = 8*4 + 7 = 39; x = 8*5 + 7 = 47; x = 8*6 + 7 = 55; etc. So, "x" can be any of these values: 15, 23, 31, 39, 47, 55, 63, 71, 79, 87, 95, 103, 111, 119, 127, 135, etc.

Clue number 6 states that when seven apples are removed at a time, no apples are left in the basket. This means "x" is a multiple of 7. From my list above, I see that the first number which is divisible by 7 is 63, and the next is 119. This is getting better; I can narrow down my search. Now, 119 - 63 = 56 (LCM of 7 and 8). Therefore, I need to keep adding 56 to get all possible values of "x" (remember, I don't need to go beyond 600). Therefore, "x" can be one of these numbers: 63, 119, 175, 231, 287, 343, 399, 455, 511, 567. Only ten numbers! That's manageable!

Clue number 5 states that when six apples are removed at a time, five apples remain in the basket. This implies that x = 6a + 5. Again, I went through the laborious process of finding "a" for all the ten values of "x":

6a + 5 = 63; so, a = 58/6. But, "a" has to be a natural number, while 58/6 is not. I can, therefore, delete 63 from my list.

Next I tried 6a + 5 = 119; I get a = 19. This is fine, and 119 remains in the list.

Next, 6a + 5 = 175; a, in this case, is not a natural number. So, 175 goes off the list.

Let's see what we get when 6a + 5 = 231; a = 37.66. It's not a natural number; 231 is erased from the list.

6a + 5 = 287; a = 47. Good, 287 stays.

6a + 5 = 343; a = 56.33. We remove 343 from the list.

6a + 5 = 399; a = 65.66. Goodbye, 399.

6a + 5 = 455; a = 75. Good, 455 stays.

6a + 5 = 511; a = 84.33. Goodbye, 511.

6a + 5 = 567; a = 93.66. Goodbye, 567.

Wonderful, I am now left with only three numbers: 119, 287, and 455.

When five apples are transferred at a time, four remain. Thus, x = 5a + 4. Just three numbers to be tried out:

5a + 4 = 119; a = 23. Fine, 119 stays.

5a + 4 = 287; a = 56.6. Goodbye, 287.

5a + 4 = 455; a = 90.2. Goodbye, 455.

Hey, that leaves me with only 119. So, 119 is the winner! Do we need to try out the remaining clues? Let's check. When four apples are transferred at a time, three remain. Therefore x = 4a + 3. 4a + 3 = 119; a = 29. That's perfect. When three apples are transferred at a time, two remain; so x = 3a + 2. 3a + 2 = 119; a = 39. Perfect, once more!. When two apples are transferred at a time, one remains; so, x = 2a + 1. This, of course, means that x is odd; 119 is certainly an odd number.

A somewhat better solution:

No, the problem did not require so many steps to be solved. It could have been solved in fewer steps.

Well, as before, let the number of apples be "x".

When I transfer two apples at a time, one remains. This implies that if I add one more apple, increasing the number of apples to (x + 1), then (x + 1) would be divisible by 2; in other words, (x + 1) will be a multiple of 2. As an example, let's say there are x = 9 apples in the basket. I transfer two apples, 7 remain; I again transfer two apples, 5 remain; I again transfer two apples, 3 remain; I do this for the fourth time, and one apple remains. Instead of 9 if I had 10 apples, no apples would have been extra.

Likewise, when I transfer three apples at a time, two remain. So, if I had (x + 1) apples, no apples would have remained; (x + 1) would be a multiple of 3. This is true for 4, 5, 6 and 8 also; (x + 1) would be divisible by 4, 5, 6 and 8. Notice that we have left out 7; when seven apples are transferred at a time, no apples remain. This means "x", and not (x + 1), is divisible by 7. (x + 1) is not a multiple of 7.

So, we have that (x + 1) is a multiple of 2, 3, 4, 5, 6 and 8. The least common multiple (LCM) of 2, 3, 4, 5, 6, and 8 is 120. Thus, 120 is divisible by 2, 3, 4, 5, 6 and 8. To find other multiples, we only have to keep adding 120: Thus, the multiples are 120, 240, 360, 480, 600, 720, 840, 960, 1080, 1200, etc.

But, I am not interested in (x + 1), I wish to know the value of x. So, the possible values of "x" are 119, 239, 359, 479, 599, 719, 839, 959, 1079, 1199, etc; of course, I don't have to consider the multiples greater than 599 because I know the basket holds at most 600 apples.

Now, 7 comes into the picture. When seven apples are transferred at a time, no apples remain; this means "x" is a multiple of 7, or in other words it is divisible by 7.

So, now all I have to do is to check which of the "x" is divisible by 7; I don't have to check beyond 599. Only 119 is divisible by 7. There were 119 apples in the basket.

Just out of curiosity let's check which would have been the next number to satisfy all the clues. It would be 959. Did I get that right?

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