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Let the 2 digit number be ab.

according to the question ab*ab*ab must be ending with 44.

The only way a cube of any number ends with the a 4 is when the number(ab's) units digit is 4.

So that narrows down the possible values of ab to 14,24,34,44,54,64,74,84,94.

Out of these 9, only 14 and 64 end with 44.

So the answer will be 14+64 = 78
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Chandanasreenivas
Let the 2 digit number be ab.

according to the question ab*ab*ab must be ending with 44.

The only way a cube of any number ends with the a 4 is when the number(ab's) units digit is 4.

So that narrows down the possible values of ab to 14,24,34,44,54,64,74,84,94.

Out of these 9, only 14 and 64 end with 44.

So the answer will be 14+64 = 78


How did you end up with 2 options out of 9?
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[quote="ParitoshRawat99"]Well first of all we can easily conclude that the last digit of the two digit number has got to be 4, as no other number would yield 4 in the unit's place after being cubed.
Now comes the tricky part.
In order to determine the second digit of the number we would keep in mind that the square of the second digit when multiplied by four should yield 4 in units digit.
Thus, possible combinations would be
1,1,4 = 4 in units digit
6,6,4 = 144, 4 in units digit

Thus the two numbers would be 14 and 64

Thus, answer = 78

Question>>>>>
"In order to determine the second digit of the number we would keep in mind that the square of the second digit when multiplied by four should yield 4 in units digit." >>> Why have you mentioned this? Is there any certain rule of this in Number properties?
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ParitoshRawat99
Well first of all we can easily conclude that the last digit of the two digit number has got to be 4, as no other number would yield 4 in the unit's place after being cubed.
Now comes the tricky part.
In order to determine the second digit of the number we would keep in mind that the square of the second digit when multiplied by four should yield 4 in units digit.
Thus, possible combinations would be
1,1,4 = 4 in units digit
6,6,4 = 144, 4 in units digit

Thus the two numbers would be 14 and 64

Thus, answer = 78

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­Can you explain im detail how you arrived at the tenth value digit?

KarishmaB Bunuel ScottTargetTestPrep Please look into the question and provide a solution.­
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Bunuel
A two digit number is formed such that the last two digits of the cube of the number is 44. What is the sum of all the possible values of the number?

(A) 72

(B) 78

(C) 80

(D) 86

(E) 98


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­
The cube of a numebr will end with a 4 only when the number ends with a 4. Recall cyclicity of units digits.

So let the two digit number be (10a+4)

\((10a+4)^3 = 1000a^3 + 64 + 120a(10a + 4) = 1000a^3 + 1200a^2 + 480a + 64 \)

Since first two terms end with 00s, only last two terms will add up to give 44 in the last two digits. 

480 * a + 64 will end with a 4 anyway in the units digit since units digits are 0 + 4 = 4

48*a should end with an 8 so that when it is added to 6, it gives 14 i.e. a 4 in the tens digit too.
So 'a' can take only two values - 1 and 6 because 48*1 ends with an 8 and 48*6 ends with an 8.

Hence the numbers must be 14 and 64 and their sum would be 78

Answer (B)
 ­
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