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too tough for me to understand

by gamtsam » Fri Aug 16, 2013 8:38 am
N is a perfect square having at least 3 digits.its last two digits are equal and not equal to zero.
how many five-digit values can N assume ?
1)8
2)6
3)7
4)9
5)5

N is a perfect square. so the last digit must be 0,1,4,5,6 or 9
sice the last two digits in N are equal, they must be
(a)00 (b)11 (c)44 (d)55 (e)66 (f)99
now if a perfect square ends in an odd digit the preceding digit must be even
so, (b),(d),(f) can be ruled out.again if a perfect square ends in 6, the
preceding digit must be an odd number so (e) is also ruled out.(a) is also
ruled out as N does not end in zero

so last two digits of N must be 44

can't really go beyond this step.
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by saurabhdhakad » Fri Aug 16, 2013 9:09 pm
Is OA = A for this ans?

Can be tried based on repeat pattern for a number whose square has last two digits same.
For eg: first 3 digit value with last two digits same would be : 144
That's 12*12 = 144
So , you may want to try other values of two digit number ending with 2.
62*62 = 3844.
Thus this pattern could be repeated for 5 digit values of N:

12,62,112,162,1112,1162,11112,11162.

This is how i tried it. Experts reply would be helpful on this.
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by Matt@VeritasPrep » Fri Aug 16, 2013 9:54 pm
I'm guessing it's 9, and here's my logic.

Any perfect squares have remainders of 0 (if they're even) or 1 (if they're odd) when divided by 4, so no perfect square ends in 11, 55, 66, or 99. As you mentioned, due to units digit properties, no perfect square ends in 22, 33, 77, or 88, so those are out. Hence we're limited to those ending in 44.

Now let's find a pattern. If the root ends in 2, we've got (10x + 2) * (10x + 2) = ...44, or 100x*x + 40x + 4 = ...44. Only the last two terms determine the 44, so x = 1, 6, 11, etc. So any positive integer of the form 50k + 12 (12, 62, 112, 162, etc) will end in 44.

If the root ends in 8, we have (10x + 8) * (10x + 8) = 100x*x + 160x + 64 = ...44, or 160x = ...80. So x = 3, 8, 13, etc. and any positive integer of the form 50k + 38 ends in ...44. So 38, 88, 138, 188, 238, etc.

Five digit perfect squares have roots between 100 and 316, so our roots are 112, 138, 162, 188, 212, 238, 262, 288, and 312.

LOVE this question, wish I'd written it.
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by sanju09 » Fri Aug 16, 2013 11:10 pm
Matt@VeritasPrep wrote:I'm guessing it's 9, and here's my logic.

Any perfect squares have remainders of 0 (if they're even) or 1 (if they're odd) when divided by 4, so no perfect square ends in 11, 55, 66, or 99. As you mentioned, due to units digit properties, no perfect square ends in 22, 33, 77, or 88, so those are out. Hence we're limited to those ending in 44.

Now let's find a pattern. If the root ends in 2, we've got (10x + 2) * (10x + 2) = ...44, or 100x*x + 40x + 4 = ...44. Only the last two terms determine the 44, so x = 1, 6, 11, etc. So any positive integer of the form 50k + 12 (12, 62, 112, 162, etc) will end in 44.

If the root ends in 8, we have (10x + 8) * (10x + 8) = 100x*x + 160x + 64 = ...44, or 160x = ...80. So x = 3, 8, 13, etc. and any positive integer of the form 50k + 38 ends in ...44. So 38, 88, 138, 188, 238, etc.

Five digit perfect squares have roots between 100 and 316, so our roots are 112, 138, 162, 188, 212, 238, 262, 288, and 312.

LOVE this question, wish I'd written it.
Nicely done Matt! I understand in the following phrases you meant "...of the form 50k + 12 when squared will end in 44, ... of the form 50k + 38 when squared ends in ...44"
"So any positive integer of the form 50k + 12 (12, 62, 112, 162, etc) will end in 44."

"and any positive integer of the form 50k + 38 ends in ...44."
It is a nice question no doubt, but I have doubt in its source. It somehow doesn't appear to me a genuine GMAT question. Anyways, taking no credit away from you, you did the true justice to this question. This punch was the master shot towards the end,
"Five digit perfect squares have roots between 100 and 316, so our roots are 112, 138, 162, 188, 212, 238, 262, 288, and 312."
Just few lines to add in no favor of GMATians...

A perfect square has span n if its very last n decimal digits are the same and non-zero. A positive integer ending in 11, 44, 99, 66, 55, is matching, respectively, to 3, 0, 3, 2, 3 (modulo 4). In view of the fact that all squares are matching to 0 or 1 (mod 4), any square with span greater than 1 must end in 44.

A paradigm is 12^2 = 144 or 38^2 = 1444.

If we read the given question carefully, we can avoid falling in trap of picking a square ending in ...444
The mind is everything. What you think you become. -Lord Buddha



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by gamtsam » Sat Aug 17, 2013 1:03 am
Thanks a ton matt,
nicely explained.
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by lunarpower » Sat Aug 17, 2013 5:29 am
Matt@VeritasPrep wrote:LOVE this question, wish I'd written it.
I like it too, but not as a GMAT question. It's not at all representative of anything that would ever show up on this test.

Math olympiad, sure. GMAT, nah.
Ron has been teaching various standardized tests for 20 years.

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by Matt@VeritasPrep » Sat Aug 17, 2013 4:01 pm
sanju09 wrote:
Nicely done Matt! I understand in the following phrases you meant "...of the form 50k + 12 when squared will end in 44, ... of the form 50k + 38 when squared ends in ...44"
"So any positive integer of the form 50k + 12 (12, 62, 112, 162, etc) will end in 44."

"and any positive integer of the form 50k + 38 ends in ...44."
Ha, yes! You hit me right in the touché - I deserved that after my quibbling elsewhere. In my defense, I'm posting remotely from an iPad this weekend and having a hard time doing the superscript squares that I like to use ... but that was sloppy. Glad it still came through.
It is a nice question no doubt, but I have doubt in its source. It somehow doesn't appear to me a genuine GMAT question.
It does seem challenging for the GMAT. Is this a CAT question, gamtsam? It is not as hard as some of the ones in Sinha's CAT math books so it seems like fair game. Curious to know the source.

In terms of American math tests, it seems like a Q16-20 level question on the AMC 12, or maybe something from the Mandelbrot.
"Five digit perfect squares have roots between 100 and 316, so our roots are 112, 138, 162, 188, 212, 238, 262, 288, and 312."
If we read the given question carefully, we can avoid falling in trap of picking a square ending in ...444
That would still be valid, though - we only need to have the last two digits be 44, so a square ending in 444 is acceptable too.
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