problem of apple?

Question

vendor who is an ex-mathematician has a number of apples and when he arranges them in rows of 3 he is left with one more, when he arranges them in rows of 5 again he is left with 1 more, Same happens with when he tries to arrange them in rows of 7 and 9 that is 1 apple is left extra.
But when he arranges them in a row of 11 he is left with none. Can u tell me how many apples were there?

Answer 1

You're looking for a multiple of 11 that, when divided by 3, 5, 7, or 9 leaves a remainder of 1. Since 3 is a factor of 9, anything with a remainder of 1 for nine rows will automatically have a remainder of 1 for three rows as well.

X = 11a = 9b + 1 = 7c + 1 = 5d + 1. Subtracting 1, we get
X - 1 = 11a - 1 = 9b = 7c = 5d.

Numbers that fit for b, c, and d are multiples of (5)(7)(9) = 315.
Now simply find a multiple of 315 that is one less than a multiple of 11.
315 + 1 = 316 / 11 = 28 8/11... nope.
630 + 1 = 631 / 11 = 57 4/11... nope.
945 + 1 = 946 / 11 = 86... ding! We have a winner.

There are 946 apples. There could also be 4411 apples, 7876 apples, or any number of apples of the form
946 + 3465k, where k is any whole number. (3465 = 315 × 11.)

Answer 2

It has to be a multiple of 11, but end in a 1 or 6.

11, 66, 121, 176, 231, 286, 341, 396, 451, 506, 561, 616, 671, 726, 781, 836, 891, 946, 1001, 1056, 1111, 1166, 1221, 1276, 1331, 1386, 1441, 1496, 1551, 1606, 1661, 1716, 1771, 1826, 1881, 1936, 1991

It cannot be divisible by 3, and the sum of the digits, when divided by 3, must leave a remainder of 1:

121, 286, 451, 616, 781, 946, 1111, 1276, 1441, 1606, 1771, 1936

It cannot be divisible by 7:

121, 286, 451, 781, 946, 1111, 1276, 1441, 1606, 1936

Testing possibilities, 946 is the only number that satisfies all the requirements. There are no doubt other numbers that also satisfy the requrements, but this is the smallest positive integer.

Answer 3

rst_ca has the right answer, just a tad quicker than me.

Easiest way is to build a spreadsheet using the multiples of 11 on the left side, then in subsequent columns do MOD 3, MOD 5, MOD 7, MOD 9 on the multiple of 11 until you find one that has a 1 for each column.

Interestingly, the first negative number is -2519.

Also, every 3465 the pattern repeats which happens to be 3 * 3 * 5 * 7 * 11 which is the product of all of the factors in the problem.

Add or subtract 3465 to 946 to get an infinite number of solutions.

Answer 4

You are looking for a number which is
1) One more than a multiple of 3=>3t+1
2)One more than a multiple of 5=>5f+1
3)One more than a multiple of 7=>7s+1
4)One more than a multiple of 9=>9n+1
5)Exactly a multiple of 11=>11e

One way to gain insight is to mix and match the various conditions to get equations. For example, 3t+1=9n+1
=>3t=9n, t=3n.

Other things to note: the number cannot have 3,5,7 as factors. Look at the numbers which have remainder=1 when divided by 3: 1, 4, 7, 10, 13...
5: 1, 6, 11, 16...
7: 1, 8, 15, 22...
9: 1, 10, 19, 28...

and the numbers divisible by 11:

0, 11, 22, 33...

You are looking for the numbers which appear in all of these lists.

Answer 5

Is he eating the apples as he arranges them in these various rows?

Seriously, it has to be some mutiple of 11. 44 almost fits the criteria, that is, 9 times 5 equals 45 which is one more than 44, however 7 does not divide into 45 so I'm flumuxed.

Answer 6

946
There may be more answers but this is the first multiple of 11 that has a remainder of 1 when divided by 3, 5, 7 and 9

Answer 7

You stated, "But when he arranges them in a row of 11 he is left with none", thus a row of 11 is 11!

Answer 8

apple

Answer 9

946..

solution is find the lcm of (3, 5, 7, 9) + 1

Answer 10

the answer may be 11 or the multiple of 11