Number stuff (urgent)

• Aug 10th 2009, 01:29 AM
justanotherperson
Number stuff (urgent)
digits 2, 3, 4, 5 and 7 are each used once to compose a 5-digit number abcde such that 4 divides a 3 digit number abc, 5 divides a 3 digit number BCD and 3 divides a 3 digit number cde. Find the 5=digit number abcde.

thnx a million
justanotherperson
• Aug 10th 2009, 02:12 AM
Gamma
72453
• Aug 10th 2009, 02:19 AM
Gamma
It was marked urgent, so I will expand on why/how I got this answer.

you know that 5 divides bcd, the only numbers divisible by 5 end in 5 or 0, 0 is not an option, so we know that d must be 5.

now move on to the fact that cde must be divisible by 3. Any number divisible by 3 must have digits which sum to something divisible by 3. we know the middle digit is 5, which leaves only a few choices of other numbers that will sum to something divisible by 3.

354
453
357
753

Now we know that 4 must divide abc, but that means for sure that abc must be even, in particular c must be even, the only one of these numbers which has c even is 453, so now we have locked in cde.

We now proceed to figure out ab4, which must be divisible by 4. Any number divisible by 4 must have the last two digits divisible by 4. In particular we only have 2 and 7 to choose from, 4 does not divide 74, and it does divide 24, so our number for abc must be 724 (you could just check to see that 274 is not divisible by 4).

This gives us the complete number, 72453 as above. Hope this helps.
• Aug 10th 2009, 03:02 AM
mathsquest
Now we know that 4 must divide abc, but that means for sure that abc must be even, in particular c must be even, the only one of these numbers which has c even is 453, so now we have locked in cde.

Shouldn't that be 354 and abc?
• Aug 10th 2009, 04:22 AM
stapel
• Aug 10th 2009, 05:43 AM
Soroban
Hello, Just!

Gamma did an excellent job!
Here's my approach to it . . .

Quote:

Digits 2, 3, 4, 5 and 7 are each used once
to compose a 5-digit number $\displaystyle abcde$ such that:
. . 4 divides the 3-digit number $\displaystyle abc$,
. . 5 divides the 3-digit number $\displaystyle bcd$, and
. . 3 divides the 3-digit number $\displaystyle cde.$

Find the 5-digit number $\displaystyle abcde.$

Since 5 divides $\displaystyle bcd$ . . . then $\displaystyle d = 5\quad\Rightarrow\quad a\;b\;c\;5\;e$
. .
We have digits: {2, 3, 4, 7}

Since 4 divides $\displaystyle abc$ . . . then $\displaystyle bc$ is a multiple of 4.
There are three choices: $\displaystyle bc \:=\:\{24, 32, 72\}$

If $\displaystyle c = 2$, we have: .$\displaystyle a\;b\;2\;5\;e$
Since 3 divides $\displaystyle cde$, then: $\displaystyle c+d+e$ is a multiple of 3.
. . This means: $\displaystyle e = 2,5,8$ . . . clearly impossible.

Hence: .$\displaystyle bc = 24 \quad\Rightarrow\quad a\;2\;4\;5\;e$
. .
We have digits: {3, 7}

Since 3 divides $\displaystyle cde$, then $\displaystyle e = 3 \quad\Rightarrow\quad a\;2\;4\;5\;3$

Therefore: .$\displaystyle 7\;2\;4\;5\;3$

• Aug 10th 2009, 11:31 AM
Gamma
Quote:

Originally Posted by mathsquest
Now we know that 4 must divide abc, but that means for sure that abc must be even, in particular c must be even, the only one of these numbers which has c even is 453, so now we have locked in cde.

Shouldn't that be 354 and abc?

I see the confusion. We were trying to lock in on cde when we determined 453. remember we got those 4 choices because 3 divides cde. But like you said we know abc must be even, fortunately abc and cde overlap in the c place, so from 4|abc we know c must be even, and from 3|cde we know in fact cde must then be 453.

Now you can proceed to see whether ab is 72 or 27. 274 is not divisible by 4, so it must be 72 for ab giving the desired.

72453