Question 1182798: The 4-digit number pqrs has the property that pqrs*4=srqp. If p=2, what is the value of the 3-digit number qrs?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
You will find nice solution/solutions to this problem under this link
https://math.stackexchange.com/questions/1839217/pqrs-cdot-4-srqp-then-what-is-the-value-of-qrs
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
In the link from the other tutor, several slightly different algebraic solutions to the problem are shown.
I think the problem is easily solved without algebra, using what we know from early in our mathematics education about multiplication.
We have...
pqrs
* 4
----
srqp
and we are told that p is 2.
Note that knowing that p=2 only makes the problem easier to solve; there is in fact only one solution to the problem, and in that solution p is 2. So let's look at the problem without using the given information that p is 2. We will quickly determine that by logical analysis.
(1) In the units column of the multiplication, we have s times 4 giving final digit p; that means p is even.
(2) For the leading digits, we have p times 4 giving us s; and the product is a 4-digit number.
So (1) and (2) tell us p is even, and 4 times p is less than 10. p can't be 0, so p can only be either 1 or 2. p "even" and "either 1 or 2" means p is 2.
So now we have...
2qrs
* 4
----
srq2
(3) Now in the units column we have s times 4 giving final digit 2, so s is either 3 or 8.
(4) And for the leading digits we have 2 times 4 giving final digit s, so s is either 8 or 9.
So (3) and (4) tell us that s must be 8.
So now we have...
2qr8
* 4
----
8rq2
(5) For the last two digits, we now have 4 times "r8" giving final digit 2. 4 times r is even, and the 4 times 8 gives us a "carry" of 3; that means q is odd.
(6) For the first two digits, we have "2q" times 4 giving us "8r"; that means q is at most 2.
So q is an odd digit that is at most 2; of course that means q is 1.
So now we have...
21r8
* 4
----
8r12
(7) Again for the final two digits we have "r8" times 4 giving us tens digit 1. With the "carry" of 3 from the units column, that means r times 4 gives us final digit 8, so r is either 2 or 7.
(8) And for the leading two digits, "21" times 4 gives us leading digits "8r", so r can be 4, 5, 6, or 7.
And that tells us r must be 7.
So now we have the unique final answer for pqrs times 4 being equal to srqp:
2178
* 4
----
8712
Solved... without algebra.
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