four digit number - good one

[helloriteshranjan]

Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?

o/1/2/3 or 4

ans is 1
and the number is 7744.
It appeared in management entrance exam. I am not sure which one. thought interesting, so posted.

[Ben Ku]

Thanks. This is an interesting problem. Your answer is right!

If we let the four-digit number be XXYY, then this number can be expressed as:
1000X + 100X + 10Y + Y = 1100X + 11Y = 11(100X + Y) = k^2 (since it's a perfect square)
In order for this to be true, 100X + Y must be the product of 11 and a perfect square, and looks like X0Y. So now our question is "which product of 11 and a perfect square looks like X0Y?" We can test them:
11 x 16 = 176; 11 x 25 = 275; 11 x 36 = 396; 11 x 49 = 593; 11 x 64 = 704; 11 x 81 = 891
The only one that fits the bill is 704. This means there is only one four-digit number that works, and it's 7744. Enjoy!

[amar.doshi]

This problem was pretty challenging... I think I found a way to solve it, but probably not something I would have been able to solve in 2 mins had I seen it for the first time! Here's how I approached it:

1 * 11 = 11
2 * 11 = 22
...
10*11 = 110
...
100*11 = 1100
101*11 = 1111
102*11 = 1122
...
401*11 = 4411
...

Hence (x0y)*11 = xxyy

for xxyy to be a perfect square, x0y itself must be equal to (11 * a perfect square)

is (11*2-square)*11 = xxyy?... 121*4 = 484, not four digits

is (11*5-square)*11 = xxyy?... 121*25 = 3025, four digits but no xxyy format

is (11*6-square)*11 = xxyy?... 121*36 = 4356, four digits but no xxyy format

is (11*7-square)*11 = xxyy?... 121*49 = 5929, four digits but no xxyy format

is (11*8-square)*11 = xxyy?... 121*64 = 7744, valid!

is (11*9-square)*11 = xxyy?... 121*81 = 9801, four digits but no xxyy format

is (11*10-square)*11 = xxyy?... 121*100 = 12100, five digits, so stop

Hence just 1 possibility

Is there a short cut based on standard math rules we are taught by MGMAT?

[RonPurewal]

This problem was pretty challenging... I think I found a way to solve it, but probably not something I would have been able to solve in 2 mins had I seen it for the first time! Here's how I approached it:

1 * 11 = 11
2 * 11 = 22
...
10*11 = 110
...
100*11 = 1100
101*11 = 1111
102*11 = 1122
...
401*11 = 4411
...

Hence (x0y)*11 = xxyy

for xxyy to be a perfect square, x0y itself must be equal to (11 * a perfect square)

is (11*2-square)*11 = xxyy?... 121*4 = 484, not four digits

is (11*5-square)*11 = xxyy?... 121*25 = 3025, four digits but no xxyy format

is (11*6-square)*11 = xxyy?... 121*36 = 4356, four digits but no xxyy format

is (11*7-square)*11 = xxyy?... 121*49 = 5929, four digits but no xxyy format

is (11*8-square)*11 = xxyy?... 121*64 = 7744, valid!

is (11*9-square)*11 = xxyy?... 121*81 = 9801, four digits but no xxyy format

is (11*10-square)*11 = xxyy?... 121*100 = 12100, five digits, so stop

Hence just 1 possibility

Is there a short cut based on standard math rules we are taught by MGMAT?

nice. notice that you basically duplicated ben ku's method, from above, with one exception: you discovered that XXYY = 11(X0Y) by random trial/error/pattern recognition, while ben found it by factoring. otherwise, your solutions seem to be exactly the same.

--

i can't really see a simpler way to solve this problem. per the original post, it was also taken from an exam that has nothing to do with the gmat - and you can certainly rest assured that you're not going to see something this obnoxious on the gmat - so you really don't have to worry about it.