[Prof. Templeton]

Choose any integer, X, between 10 and 1000. Remove its last digit to get another integer, Y. (For example, if X is 356, Y is 35.) What is the probability that X is evenly divisible by Y?

[unreality]

are 10 and 1000 included in the random range, or is it 11-999 or 11-1000 or 10-999 ?

[Prof. Templeton]

are 10 and 1000 included in the random range, or is it 11-999 or 11-1000 or 10-999 ?

They are included

[unreality]

my attempt:

the number's digits are a, b, and then Z. abZ for 2,3 and 4 means the number 234, not 2*3*4

X = abZ

Y = ab

Z = Z

for the last digit, Z, 0 happens 100/991 times and each of the other 9 digits happens 99/991 times

it is not evenly divisible if the last digit (Z) is 0, so every multiple of 10 is not allowed. So 100 out of the 991 numbers are NO.

What if Z is 5? Y would have to end in 0 or 5, ie b would have to be 0 or 5, which there is a 1/5 chance that b is 0 or 5 if Z is 5. 1/5 of 99 is 19.8. The other 4/5, or 79.2, is NO. So far we have 179.2/991 as NO

moving on... what if Z is 2? Y would have to end in an even number, which would be slightly more for X since 10 and 1000 are included, but Y is different and is thus 1/2. So 1/2 of 2's 99 cut that it's a NO, making the total so far 228.7/991 of it not being possible, and so far we've covered 0, 5 and 2 for X.

am i on the right track, Prof Templeton?

[Guest]

Alright, first shot:

Ok, first to find the pattern

for 10-19: 100%

for 20-29: 50%

for 30-39: 40%

for 40-49: 30%

for 50-59: 20%

for 60-69: 20%

for 70-79: 20%

for 80-89: 20%

for 90-99: 20%

for 100-109: 10%

for 110-119: 10%

for 120-129: 10%

for 130-139: 10%

...

for 190-199: 10%

...

for 570-579: 10%

...

for 990-999: 10%

for 1000: 100%

-------------------------------------------------

Since this is a decaying pattern:

10-19: 100% all numbers are dividable by 1

20-29: 50% all evens (20, 22, 24, 26, 28)

30-39: 40% all thirds (30, 33, 36, 39)

40-49: 30% all 4ths (40, 44, 48)

50-99: 20% the base number and one other (50, 55, 60, 66, 70, 77... )

100-999: 10% only the base number will work (100, 110, 120, 130 ... )

1000: 100% no duh

------------------------------------

which makes:

((10*1) + (10*.5) + (10*.4) + (10*.3) + (50*.2) + (900*.1) + (1*1))/990 = .1242 or 12.42%

EDIT: I think I took the long way around

Edited August 8, 2008 by pw0nzd

[Guest]

my attempt:

the number's digits are a, b, and then Z. abZ for 2,3 and 4 means the number 234, not 2*3*4

X = abZ

Y = ab

Z = Z

for the last digit, Z, 0 happens 100/991 times and each of the other 9 digits happens 99/991 times

it is not evenly divisible if the last digit (Z) is 0, so every multiple of 10 is not allowed. So 100 out of the 991 numbers are NO.

What if Z is 5? Y would have to end in 0 or 5, ie b would have to be 0 or 5, which there is a 1/5 chance that b is 0 or 5 if Z is 5. 1/5 of 99 is 19.8. The other 4/5, or 79.2, is NO. So far we have 179.2/991 as NO

moving on... what if Z is 2? Y would have to end in an even number, which would be slightly more for X since 10 and 1000 are included, but Y is different and is thus 1/2. So 1/2 of 2's 99 cut that it's a NO, making the total so far 228.7/991 of it not being possible, and so far we've covered 0, 5 and 2 for X.

am i on the right track, Prof Templeton?

Just a quick look at this so far, but

numbers ending in zero a NO for you? I'm pretty sure 770/77 = 10 which fits the requirement of the post.

[Prime]

The first answer that jumps to mind is 100/991.

[Prof. Templeton]

The probability that a randomly chosen integer X, between 10 and 1000, will be divisible by a second integer Y, formed by all but the last digit of X, is 123/991 (almost 1 in 8).

There are 991 possible values for X (1000 - 10 + 1 = 991). Those that end with 0 will yield a Y that exactly divides X, yielding a quotient of 10 (for example, 230 / 23 = 10). There are 100 such "good" values for X.

If the last digit of X is not zero, the result of dividing X by Y will be a quotient greater than 10. For example, if X is 22, Y is 2, and the quotient is 11. All two-digit multiples of 11 (11, 22, 33, 44, etc.) have this property; there are 9 such multiples.

If X contains more than two digits, it is easy to see that X/Y must be less than 11 (the maximum quotient in this case is 10.9, corresponding to X = 109). Hence there can be no "good" values for X having three or more digits except for the multiples of 10 already counted. It thus remains necessary only to consider two-digit values for X.

In addition to the first 9 multiples of 11, the first 5 multiples of 12 (12, 24, 36, 48, and 60) are evenly divisible by their first digits (but 60 has already been counted, since it ends in 0). Similarly, the first 3 multiples of 13 (13, 26, and 39), and the first 2 multiples of 14 (14 and 28) are "good" values. Finally, the 5 remaining two-digit numbers beginning with 1 (15, 16, 17, 18, and 19), but not their multiples, are "good" numbers. In all, there are 100 + 9 + 4 + 3 + 2 + 5 = 123 "good" values among the 991 possible values of X, so the probability of selecting a "good" X at random is 123/991, or about 12.4%.

[Guest]

Ah, so the pattern does in fact continue past 1000. Great puzzle, Prof. Templeton.

[Prof. Templeton]

Ah, so the pattern does in fact continue past 1000. Great puzzle, Prof. Templeton.

Kudos to you for solving so quickly.

[Guest]

"Choose any integer, X, between 10 and 1000. Remove its last digit to get another integer, Y. (For example, if X is 356, Y is 35.) What is the probability that X is evenly divisible by Y? "

Technically the probability is either 1 or zero depending on what number is chosen. It is either evenly divisible 2nd number or not.

But to solve the intended question

10-19 divisible by 1 =10/10

20-29 by 2 = 5/10 20 22 24 26 28

30-39 by 3 =4/10

40-49 3/10

50-99 2/10 *5

from 100- 1000 only multiples of 10 so 91/1000

so 10+ 5+4+3+10+91=123

so 123/990= .1242424

[Guest]

"Choose any integer, X, between 10 and 1000. Remove its last digit to get another integer, Y. (For example, if X is 356, Y is 35.) What is the probability that X is evenly divisible by Y? "

I understand the spirit of the question. I solved it and got

the 123/991 but I think the actual answer to the question is 100% or 0% depending on what number "You" choose. It does not ask for the cumulative probability only the probability of the chosen integer. Ex if I pick 356 the prob is 0%

[unreality]

I misunderstood the question. What I thought it meant was this:

you pick 557. Is 55/7 a whole number? (answer: no)

Which is why things ending in 0 are not allowed, ie:

you pick 320. 32/0 is not a whole number

etc

[Prime]

The probability that a randomly chosen integer X, between 10 and 1000, will be divisible by a second integer Y, formed by all but the last digit of X, is 123/991 (almost 1 in 8).

There are 991 possible values for X (1000 - 10 + 1 = 991). Those that end with 0 will yield a Y that exactly divides X, yielding a quotient of 10 (for example, 230 / 23 = 10). There are 100 such "good" values for X.

If the last digit of X is not zero, the result of dividing X by Y will be a quotient greater than 10. For example, if X is 22, Y is 2, and the quotient is 11. All two-digit multiples of 11 (11, 22, 33, 44, etc.) have this property; there are 9 such multiples.

If X contains more than two digits, it is easy to see that X/Y must be less than 11 (the maximum quotient in this case is 10.9, corresponding to X = 109). Hence there can be no "good" values for X having three or more digits except for the multiples of 10 already counted. It thus remains necessary only to consider two-digit values for X.

In addition to the first 9 multiples of 11, the first 5 multiples of 12 (12, 24, 36, 48, and 60) are evenly divisible by their first digits (but 60 has already been counted, since it ends in 0). Similarly, the first 3 multiples of 13 (13, 26, and 39), and the first 2 multiples of 14 (14 and 28) are "good" values. Finally, the 5 remaining two-digit numbers beginning with 1 (15, 16, 17, 18, and 19), but not their multiples, are "good" numbers. In all, there are 100 + 9 + 4 + 3 + 2 + 5 = 123 "good" values among the 991 possible values of X, so the probability of selecting a "good" X at random is 123/991, or about 12.4%.

But of course! I forgot to take those into account. Nice trick, Prof. T!

[Guest]

Choose any integer, X, between 10 and 1000. Remove its last digit to get another integer, Y. (For example, if X is 356, Y is 35.) What is the probability that X is evenly divisible by Y?

Let X = 100x + 10y + z and Y = 10x + y

Then synthetic division gives X/Y = 10 + z/(10x + y)

For X/Y to be an integer:

1. z = 0 and that gives 100 possibilities

2. x = 0 and z & y can take on 1 thru 9 and z/y an integer 9 + 4 +3 + 2 + 1 + 1 + 1 + 1 + 1 = 23 more possibilities

Therefore 123 out of 991 or .1241