what is the smallest whole number which,?

Question

what is the smallest whole number which,
when divided by 10 leaves a remainder of 9,
when divided by 9 leaves a remainder of 8,
when divided by 8 leaves a remainder of 7,
when divided by 7 leaves a remainder of 6,
when divided by 6 leaves a remainder of 5,
when divided by 5 leaves a remeinder of 4,
when divided by 4 leaves a remainder of 3,
when divided by 3 leaves a remainder of 2, and
when divided by 2 leaves a remainder of 1?

Answer 1

1

divide by 10 is 0 r 9 (9+1 =10)
divide by 9 is 0 r 8 (8+1 = 9)
divide by 8 is 0 r 7 (7+1 = 8)
etc etc etc. . .

Answer 2

To figure out this number you can use a factor tree. You first start with the roots and work your way up. Remember that in a factor tree you must break up every number into the smallest possible prime number example 45 l ___ l 9 __ 5 l l 3 3 So 45 is just 3*3*5 it is also divisble by any combination of those numbers like 9*5 or 15*3 or 45*1 So in our roots we must have a 2 3 another 2 to make it divisble by 4 since 2*2 = 4 5 we don't need do add another number for six since we already have a 3 and 2 (3*2=6) 7 another two to make 8 since 2*2*2=8 and last we need another 3 since 3*3 = 9 so we have 2*3*2*5*7*2*3= 2520 you can combine any of those numbers to make 2,3,4,5,6,7,8, and 9

Answer 3

Notice that if you added one to this number, it would be divisible by 10, 9, 8, 7, 6, 5, 4, 3, and 2. So the answer will be the smallest number divisible by all of these, minus one.

The smallest number divisible by the numbers 2 through 10 is 2^3 x 3^2 x 5 x 7 = 2520.

So the answer to your question is 2519.

Answer 4

9
8
7
...
1

That is, when you divide 9 by 10, you get zero with a remainder of 9
The same logic holds to all of these.