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Problem 1 : An integer greater than 1 that is not a prime is

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Problem 1 : An integer greater than 1 that is not a prime is [#permalink]

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12 Jun 2006, 15:41
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Problem 1 : An integer greater than 1 that is not a prime is called composite.
If the two digit integer n is greater than 20, is n composite ?

1. The tenths digit of n is a factor of the unit digits of n
2. The tenths digit of n is 2

Solution

from the statement we know n is between 21-99

from (1)
we get that n can be 2x, 3x, 4x

so the numbers can be 22, 24, 26, 28, 33, 36,39, 44, 48

so (1) is sufficient

from (2) not enough could be 23 or 24 or whatever

so I'll go with A on this one
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12 Jun 2006, 15:54

How many integers between 324,700 and 458,600 have tens digit 2 and units digit 1?

1,339
1,352
1,353
10,030
10,300

458,600-324,700 =133,900

tens digit 2 and units digit 1 this mean 21, this number will appear once in every 100.

133900/100 = 1339
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12 Jun 2006, 16:07
Problem 3: What 2 digit number (which is a integer) is the product of its digits. I have worked this problem out but looking for other solutions.

Solution1:Let x be the digit in ten's place and y be the digit in unit's place.
From question stem: 10x + y = 2xy
Therefore, x = y/2(y-5)
Since x and y are digits of a 2-digit number, x and y are positive integers.
So, y must be greater than 5 and also y must be even.

Try the first even number greater than 5 => 6
If y = 6, x = 3 --> 2-digit number = 36

Solution 2 : The way i got it is :
2 digit number=> xy
Now as per the stem 10x+y=2(x*y)
10x+y=2xy => y= (10x)/(2x-1)
Lets do plugin
x y
1 10(Not possible as no wont be 2 digit)
2 Non Integer
3 6
So 36
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12 Jun 2006, 16:16
Problem 4 : p is a positive integer. What is the remainder when 19^p is divided by 10?

1) p is a multiple of 5
2) p is a multiple of 6

B it is...

lets just focus on the unit digit of 19^p

9^0=1, 9^5=1 (i am just looking for unit digit)

i)since p is a multiple of 5, p can be 10, 15 etc.... therefor if unit digit 1 or 9, the remainder will be diff...ins (look for 9^even or 9^odd)!

ii) since a multiple of 6 is always even, 9^even unit digit is always 1, therefore we can now the remainder for 10 for sure...sufficient!
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12 Jun 2006, 16:21
What is hundredths digit of decimal z?

(1) The tenths digit of 100z is 2
(2) The units digit of 1000z is 2

z = 0.xyabc.... We need to find the value of y!!!

1) 100z = xy.abc. Tenths digit = a = 2. Insuff
2) 1000z = xya.bc. Units digit = a = 2. Insuff

Taking it together we do not have any additional info to find value of y.

Hence E.
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12 Jun 2006, 16:28
Just as an FYI :

the idea is to post questions that you have answered/feel are good Qs/ones you cannot answer, *WITHOUT* posting the answer along with the Qs.

That way it gives people the opportunity to solve and explain their logic.

Thanks.
12 Jun 2006, 16:28
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Problem 1 : An integer greater than 1 that is not a prime is

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