# 4th grade math problem

1. Oct 5, 2006

### go2cnavy

My daughter just came home with this problem and I feel really stupid cause I have looked it over for an hour and cannot get it. I just finished a calc II refresher course and maybe I am overthinking(nuking) the problem. Sure, I can randomly plug and chug some numbers through my calculator, but was wondering if someone would provide a little insight so I can explain to her how to do it?

In this puzzle, three digits are provided as clues. Fill in the boxes(X) [DECIMALS ARE ONLY PLACE HOLDERS TO GET THE CORRECT FORMATTING] below with the prime numbers, 2,3,5,or 7 so the multiplication makes sense.

Please forgive me for the poor formatting of the equation

..XXX
....XX
EQUALS
.XXXX
2XXX.
EQUALS
2X5XX

Last edited: Oct 5, 2006
2. Oct 5, 2006

### CRGreathouse

Of those digits, only 5 can appear as the units digit. That gives you two Xs right off the bat.

3. Oct 5, 2006

### go2cnavy

I don't understand the comment "units" digit. PLease clarify if you have time

4. Oct 5, 2006

### Office_Shredder

Staff Emeritus
If you have a number:

abcd

where a, b, c and d are the digits, then d is the units digit, c is the tens digit, b is the hundreds digit, etc.

you just need a 5 in the units digit of each solution, and one of the two initial units

5. Oct 5, 2006

### go2cnavy

why can 5 be the only units digit in each solution

6. Oct 6, 2006

### arildno

I misunderstood your notation, sorry.
All deleted.
I'll get back with a better reply.

7. Oct 6, 2006

### go2cnavy

I am sorry for the poor formatting.

There is to be only one multiplication involved. the subsequent X's below the first(equals) are the products of the multiplication with the 5 digit final result being the answer.

8. Oct 6, 2006

### arildno

Here's how you should start, gocnavy:
Get a better notation than the one in the book first!!
So I'll call the five numbers you are to find A,B,C,D and E.
We have:
$$A=A_{h}*100+A_{t}*10+A_{u}*1$$
$$B=B_{t}*10+B_{u}*1$$
$$C=C_{th}*1000+C_{h}*100+C_{t}*10+C_{u}*1$$
$$D=2*1000+D_{h}*100+D_{t}*10+D_{u}*1$$
$$E=2*10000+E_{th}*1000+5*100+E_{t}*10+E_{u}*1$$
we have the relations:
$$A*B_{u}=C$$
$$A*B_{t}=D$$
$$D*10+C=E$$

The indices th, h,t,u stands for "no. of thousands", "no. of hundreds", "no. of tens" and "no. of units", respectively,
and the digits $$A_{h},A_{t}, A_{u},B_{t},B_{u},C_{th},C_{h},C_{t},C_{u},D_{h},D_{t},D_{u},E_{th}, E_{t},E_{u}$$
can only be 2,3,5 or 7

Got that?

9. Oct 6, 2006

### arildno

I have seen your reply; what I've written in the previous post is concerned with how you meant the problem to be understood.

10. Oct 6, 2006

### arildno

Now, we have that $E_{u}=C_{u}$
and we must also have:
$$E_{u}=C_{u}=D_{u}=5$$
The reason for this, is that the only PRODUCTS of 2,3,5,7 that has as its last digits 2,3,5 or 7 is 3*5, 5*3, 5*5 and 7*5, all of whom having last digit 5.

Thus, we have the schema (in your notation):
..XXX A
....XX B
-----
.XXX5 C
2XX5. D
------
2X5X5 E
Looking at the next last column, we see that we must have:
a) $$(C_{t},E_{t})=(2,7)$$
or:
b)$$(C_{t},E_{t})=(7,2)$$

Last edited: Oct 6, 2006
11. Oct 6, 2006

### go2cnavy

I understand this. Thank you very much. BUT is my 4th grade daughter expected to follow this same format? Or do you see this as a plug and chug method of a math game to stumble upon the correct answer? This seems far to complicated to expect a 4th grader to do. Is there possibly a more elementary approach to the problem? I sent her teacher a npte, and her reply was the 3digit number begins with a 7 and the two digit number starts with 3.

12. Oct 6, 2006

### arildno

I am horrified that this type of exercise is given to a 4th grader (I'm not sure, but that's age group 10-11, right?).

If the kid is to do this METHODICALLY, and systematically, then this is way too cumbersome for the average kid. That is the way I wanted to show you, but I wanted some response before proceeding.

I assume this is a meant as a guess-and-play exercise, that is something wholly without pedagogical value that only will be fun for those few A+ kids who like to play around with numbers.

In my opinion, kids ought to master prescribed algorithms and develop understanding of what they are doing, they are simply not competent enough in math to find "clever shortcuts" or new ways to do math.

13. Oct 6, 2006

### arildno

Okay, now that I've cleared out a few points, we must now look in detail on the multiplication of A with $B_{t}$ (producing C) and $B_{u}$ (producing D), respectively. Since the arguments will walk along the same lines for both numbers, I'll work with a "b" that can be 2,3,5 or 7.

We therefore have:
$$A*b=A_{h}*b*100+A_{t}*b*10+A_{u}*b (EQ 2)$$
This must be a 4-digit number ending as I've said in 5, and thus b=3,5 or 7.
We will treat the cases b=7 and b=3 first, since these forces $A_{u}=5$

b=7:
We insert b=7 in (EQ 2) and rearrange it:
$$C=A*7=7A_{h}*100+(7A_{t}+3)*10+5*1$$
Thus, if b=7, then we must have $A_{t}=2 or 7$ by looking at the 10's place.
b=7,At=7:
We now have:
$$C=(7A_{h}+5)*100+2*10+5*1$$
Note that for NO choice for Ah will we get a proper 100-digit for C!
Thus, if b=7, then At cannot be 7, but must be 2!
b=7, At=2:
We get, by inserting these in (EQ 2) and rearranging:
$$C=(7A_{h}+1)*100+7*10+5*1$$
Here, we see that only values 2 and 3 are valid for Ah.

Thus, consistent with b=7, we have the possibilities:
A=225, C=1575 and A=325, C=2275
(Note that if $B_{t}=7$, then A MUST be 325, since the first digit in D is 2!)

Now, we may continue..(sigh)

14. Oct 6, 2006

### arildno

Now, note that if B=77 (and therefore A=325),
then:
.2275
2275
------
25025,
that is, not both the digits of B can be 7.
Let us therefore look at the two other cases with Bt=7, namely B=73 and 75, and A=325
B=73 can be dismissed outright, since 3*325=975, which cannot work for C.
B=75 doesn't work either, for that yields C=1625

Thus we have shown that Bt must either be 3 or 5, it cannot be 7!

15. Oct 6, 2006

### go2cnavy

wow...my daughter is 9 and still has some difficluty doing long division. She does not enjoy math enough yet for a problem like this. Her attention span with respect to mathematics is very minimal.

16. Oct 6, 2006

### arildno

Quite as I thought, a perfectly normal 9 year old.

As you can see, I am systematically going through the options, there really isn't anything to learn from such exercises, they are just tests for your endurance of numbers.

17. Oct 6, 2006

### arildno

As an aside, she really ought to be given time by the teachers to work on those long divisions, that is important stuff, whereas the given exercise is both tedious and unimportant.

18. Oct 6, 2006

### go2cnavy

I agree. She is in Venture(gifted) classes on Wednesdays and is out of her normal class room. Her primary teacher gave her this on Wednesday as home work because she missed class. She is a brilliant kid, but her gift is more for the arts. She taught herself to read by the age of three. Mathematics does not come as easy for her. Thank you for your time and effort to explain the procedure for tackling such a problem. I can see where it may draw on the interest of a child who enjoyed playing with numbers. Do we have a final answer yet?

19. Oct 7, 2006

### arildno

Give me some time. It is rather boring to do such an exercise and type it in, so I must take a break from time to time.

20. Oct 7, 2006

### go2cnavy

I understand and did not intend to rush you.

21. Oct 7, 2006

### arildno

Now, we can also readily show that $B_{u}$ cannot be 7 either (i.e B=37 or 57). Since A in that case must be either 225 or 325, multiplying either of these two numbers with 3 or 5 won't get us 2 in D's first digit.

Thus, the only four numbers possible for B is 33,35,53 and 55.

22. Oct 7, 2006

### arildno

Now, assuming one of B's digits, called b, b=3, we will find that A has to be 775, in order to fullfill all criteria (given by what C and D must be).

And indeed, with A=775, B=33, we get:
..775
...33
-----
.2325
2325
------
25575

That is, A=775, and B=33 is a solution of the exercise

Now, 775*5=3875, so this can't be either C and D. Thus, we have eliminated 35 and 53 as options.

We are left, then, with B=55.
I'll deal with that later.

23. Oct 7, 2006

### arildno

55 didn't work out, either (you may find that out on your own).

Thus, there is only one solution: A=775, B=33

24. Oct 7, 2006

### arildno

Now, I have thought of a way how you can show your daughter how to do this in way somewhat less cumbersome.

This includes some clever thinking at the start, and thus difficult for most kids to find out of their own.

First:
The biggest 3-digit number out of 2,3, 5 or 7 we can make is 777.

Now, the least 2-digit number we can form is 22.

If we multiply 22 with 777, we find that the answer is too little for D (it will have a 1 as first digit), that is, we CAN'T have 2 as our first digit in B, because any 3-digit number less than 777 will face the same problem.

Thus, we can try out multiplying 777 with 32, 33 and 37 to see if we get anything, further than that, we need not go, since 5*7=35, which gives a too big number for D.

We find that 777 can't work as our number A, so we go on to 775.
We know from previous that 2 can't be our first digit in B, so we start out multiplying 775 with 32, which doesn't work.

However, on the very next try, we are very lucky because 775*33 DOES work, and you can tell your daughter she is finished!
If she asks if there can't be any other solutions, then tell her that to answer that question with a NO, you'll need to check out all the other possibilities, and that takes a lot of time.
I'm sure she'll be happy to have found the answer, and I assume that this was the method meant that the kids should come up with on their own.

Note that in general, this technique isn't any less cumbersome than what I've done, it is only because the true solution in this particular problem happens to come so early on in the algorithm that it seems more efficient.

All algorithms that solves such problems are really to go through each option in a systematic manner. One algorithm goes through the options in its own way, another algorithm through the same options in ITS own way (ckecking the options in some other sequence).
It is mere coincidence whenever one algorithm was lucky enough to start close to the solution.

In many problems of this sort, there will be more than one solution, and thus you'll need to go through each and every option. This shows that the particular sequence you are checking your options in isn't inherently any better than any other sequence.

Note in particular that if we had happened to start out with A=222 and B=77 instead, we would have been very unlucky in our choice of algorithm for this particular problem, and there are no fundamental LOGICAL reasons that makes this unlucky choice any worse than starting out with A=777 B=22.
Thus, the unfortunate kid might start out with this choice, and simply get too tired to finish the problem, even though his reasoning behind his idea is as good as anyone else's.

That is why this is a very bad exercise.

As a final comment on my own algorithm, note that I essentially checked CLASSES of numbers with some common property (say, those possible B's having a 7 in them), rather than individual checks.
Usually, this is way faster than to try out one multiplication after the other, but it is definitely too abstract to be hit upon as a strategy by a 9-year old.

Last edited: Oct 7, 2006
25. Oct 7, 2006

### go2cnavy

Thank you very much for all of your time and suggestions. I hope it wasn't too boring for you.