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If the tens digit of the positive integer x and the tens digit of the 03 Jan 2024, 06:46
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If the tens digit of the positive integer x and the tens digit of the positive integer y are both 6, what is the number of different possible values for the tens digit of 2x + 2y?

A. Three
B. Four
C. Five
D. Six
E. Seven

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If the tens digit of the positive integer x and the tens digit of the 03 Jan 2024, 08:00
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Bunuel
If the tens digit of the positive integer x and the tens digit of the positive integer y are both 6, what is the number of different possible values for the tens digit of 2x + 2y?

A. Three
B. Four
C. Five
D. Six
E. Seven

Show Spoiler
Attachment:
Screenshot 2024-01-02 202125.png
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The tens digit of 2x+2y will depend only on the units digit, so it doesn’t matter what other digits are.
Let the two numbers be 6a and 6b.
Least value of 2(x+y) = 2(60+60) = 240
Maximum value of 2(x+y) = 2(69+69) = 276

So 2(x+y) can take all even values from 240 to 276.
Thus, tens digit can be 4, 5, 6 and 7, a total of four values.

B
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If the tens digit of the positive integer x and the tens digit of the 03 Jan 2024, 22:50
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Bunuel
If the tens digit of the positive integer x and the tens digit of the positive integer y are both 6, what is the number of different possible values for the tens digit of 2x + 2y?

A. Three
B. Four
C. Five
D. Six
E. Seven

Show Spoiler
Attachment:
Screenshot 2024-01-02 202125.png
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\(x = 6p\) ⇒ \(60 + p\)

\(y = 6q\) ⇒ \(60 + q\)

\(2x = 120 + 2p\)

\(2y = 120 + 2q\)

\(2x + 2y = 240 + 2(p+q)\)

The minimum value of \(p + q = 0\), and the maximum value of \(p + q = 18\)

Hence, the minimum value of \(2(p+q) = 0\), the maximum value of \(2(p+q) = 36\). From this information, we can infer that while performing the addition operation we can have a carryover of 0, 1, 2, or 3 from the unit place to the tens place.

Possible values of the tens place = (4+0), (4+1), (4+2), (4+3) = 4, 5, 6, or 7.

Option B
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If the tens digit of the positive integer x and the tens digit of the 05 Jan 2024, 04:57
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Bunuel
If the tens digit of the positive integer x and the tens digit of the positive integer y are both 6, what is the number of different possible values for the tens digit of 2x + 2y?

A. Three
B. Four
C. Five
D. Six
E. Seven

Show Spoiler
Attachment:
Screenshot 2024-01-02 202125.png

\(x = 6p\) ⇒ \(60 + p\)

\(y = 6q\) ⇒ \(60 + q\)

\(2x = 120 + 2p\)

\(2y = 120 + 2q\)

\(2x + 2y = 240 + 2(p+q)\)

The minimum value of \(p + q = 0\), and the maximum value of \(p + q = 18\)

Hence, the minimum value of \(2(p+q) = 0\), the maximum value of \(2(p+q) = 36\). From this information, we can infer that while performing the addition operation we can have a carryover of 0, 1, 2, or 3 from the unit place to the tens place.

Possible values of the tens place = (4+0), (4+1), (4+2), (4+3) = 4, 5, 6, or 7.

Option B
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and the maximum value of p+q=18
why's this the case please?
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If the tens digit of the positive integer x and the tens digit of the 05 Jan 2024, 05:45
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tickledpink001

and the maximum value of p+q=18
why's this the case please?
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tickledpink001 - \(p\) and \(q\) are single digits non-negative integers (i.e. both \(p\) and \(q\) can take values from \(0\) to \(9\), inclusive). Hence, the maximum value that \(p\) and \(q\) can have is \(9\), resulting in the maximum value of \(p + q = 18\).

Hope this helps.
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If the tens digit of the positive integer x and the tens digit of the 15 Mar 2024, 20:07
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We have to find the unique tens digit number from 2x + 2y

Min number could be 61 and max number could be 69.

So the range of numbers from 240 to 276 which we have to find the unique number of tens digits.

Only unique digits we will get from range 240 to 276 is 4 , 5, 6, 7 as unique tens digits

So total 4 unique digits.

Option - B will be the answer
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If the tens digit of the positive integer x and the tens digit of the 27 Apr 2024, 13:55
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Bunuel
If the tens digit of the positive integer x and the tens digit of the positive integer y are both 6, what is the number of different possible values for the tens digit of 2x + 2y?

A. Three
B. Four
C. Five
D. Six
E. Seven

Show Spoiler
Attachment:
Screenshot 2024-01-02 202125.png
The tens digit of 2x+2y will depend only on the units digit, so it doesn’t matter what other digits are.
Let the two numbers be 6a and 6b.
Least value of 2(x+y) = 2(60+60) = 240
Maximum value of 2(x+y) = 2(69+69) = 276

So 2(x+y) can take all even values from 240 to 276.
Thus, tens digit can be 4, 5, 6 and 7, a total of four values.

B
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­Why did you assume the integer is a 2 digit integer? Thank you.
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If the tens digit of the positive integer x and the tens digit of the 27 Apr 2024, 19:37
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Engineer1

chetan2u

Bunuel
If the tens digit of the positive integer x and the tens digit of the positive integer y are both 6, what is the number of different possible values for the tens digit of 2x + 2y?

A. Three
B. Four
C. Five
D. Six
E. Seven

Show Spoiler
Attachment:
Screenshot 2024-01-02 202125.png
The tens digit of 2x+2y will depend only on the units digit, so it doesn’t matter what other digits are.
Let the two numbers be 6a and 6b.
Least value of 2(x+y) = 2(60+60) = 240
Maximum value of 2(x+y) = 2(69+69) = 276

So 2(x+y) can take all even values from 240 to 276.
Thus, tens digit can be 4, 5, 6 and 7, a total of four values.

B
­Why did you assume the integer is a 2 digit integer? Thank you.
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­We don't know the number of digits, but it doesn't matter. When we add, no digit affects the value of those to its right, only those to its left. 
For instance, if we add 2169 + 53,869, the answer still ends in 38, just as when we add 69 + 69. Similarly, when we double the value, those additional places to the left don't change the fact that 38 doubles to 76. 

So in short, we can focus only on the relevant digits and not worry about any other potential digits. This is common on place value/digit problems.
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If the tens digit of the positive integer x and the tens digit of the 28 Apr 2024, 07:36
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DmitryFarber
­We don't know the number of digits, but it doesn't matter. When we add, no digit affects the value of those to its right, only those to its left. 
For instance, if we add 2169 + 53,869, the answer still ends in 38, just as when we add 69 + 69. Similarly, when we double the value, those additional places to the left don't change the fact that 38 doubles to 76. 

So in short, we can focus only on the relevant digits and not worry about any other potential digits. This is common on place value/digit problems.
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­Understood, thank you. I think you meant this, correct?  "When we add, no digit affects the value of those to its left, only those to its right."­
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If the tens digit of the positive integer x and the tens digit of the 28 Apr 2024, 18:54
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Bunuel
If the tens digit of the positive integer x and the tens digit of the positive integer y are both 6, what is the number of different possible values for the tens digit of 2x + 2y?

A. Three
B. Four
C. Five
D. Six
E. Seven

Show Spoiler
Attachment:
Screenshot 2024-01-02 202125.png
The tens digit of 2x+2y will depend only on the units digit, so it doesn’t matter what other digits are.
Let the two numbers be 6a and 6b.
Least value of 2(x+y) = 2(60+60) = 240
Maximum value of 2(x+y) = 2(69+69) = 276

So 2(x+y) can take all even values from 240 to 276.
Thus, tens digit can be 4, 5, 6 and 7, a total of four values.

B
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­What do you mean by even values? Also  why is the assumption that the "positive" integer for x and y is only 2 digits (and not bigger)? Help is appreciated :)
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If the tens digit of the positive integer x and the tens digit of the 28 Apr 2024, 19:07
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chetan2u

Bunuel
If the tens digit of the positive integer x and the tens digit of the positive integer y are both 6, what is the number of different possible values for the tens digit of 2x + 2y?

A. Three
B. Four
C. Five
D. Six
E. Seven

Show Spoiler
Attachment:
Screenshot 2024-01-02 202125.png
The tens digit of 2x+2y will depend only on the units digit, so it doesn’t matter what other digits are.
Let the two numbers be 6a and 6b.
Least value of 2(x+y) = 2(60+60) = 240
Maximum value of 2(x+y) = 2(69+69) = 276

So 2(x+y) can take all even values from 240 to 276.
Thus, tens digit can be 4, 5, 6 and 7, a total of four values.

B
­What do you mean by even values? Also  why is the assumption that the "positive" integer for x and y is only 2 digits (and not bigger)? Help is appreciated :)
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Even values meant even integers between 240 and 276.

Next if the numbers are 768964 and 876543865, we are looking for TENS digit of 2(768964 + 876543865). TENS digit will be given only by ones digit and tens digit, so why waste time on thinking what other digits are => 2(64+65)

Posted from my mobile device
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If the tens digit of the positive integer x and the tens digit of the 23 Jun 2024, 03:53
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Bunuel
If the tens digit of the positive integer x and the tens digit of the positive integer y are both 6, what is the number of different possible values for the tens digit of 2x + 2y?

A. Three
B. Four
C. Five
D. Six
E. Seven

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­x and y both are of the form ...60 or ...61 or ... or ...69 

When we add them (x + y), the sum ranges from
...60 + ...60 = ...20
...60 + ...61 = ...21
to
...68 + ...69 = ...37
...69 + ...69 = ...38

Last two digits will be anything from 20 to 38. 
When we multiply this sum by 2 to get 2(x+y), the value ranges from 

2 * ...20 = ...40
2* ...21 = ...42
to
2*...37 = ...74
2*...38 = ...76

Hence the tens digit of 2(x+y) can be anything from 4 to 7 i.e. it can take 4 distinct values. 

Answer (B)
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In both the cases when we multiply both X and Y individually 2 , the tens digit of 2X and 2Y can either be 2 or 3. Which means that the sum of tens digits of 2X and 2Y can be 4 (2+2) or 5 (2+3 , 3+2) or 6 (3+3).

Now on top of this just consider the maximum and minimum values the unit digit of (2X+2Y) can be which can be any even number between 0 to 16 (units of 2X or 2Y can be 0,2,4,6,8) . As long as the sum total of the unit digits is among 0,2,4,6,8 nothing needs to be added to the sum of the tens digit and it stays among 4 , 5 ,6 but when the sum total of unit digit is among 10, 12,14 ,16 then an additional 1 needs to be added to the tens digit and after adding 1 the tens digit can be 5(4+1) or 6(5+1) or 7(6+1) . So the probable values of tens digit can be 4 or 5 or 6 or 7 so a total of Four values and that's the answer
Bunuel
If the tens digit of the positive integer x and the tens digit of the positive integer y are both 6, what is the number of different possible values for the tens digit of 2x + 2y?

A. Three
B. Four
C. Five
D. Six
E. Seven

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If the tens digit of the positive integer x and the tens digit of the 13 Jul 2025, 10:54
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In both the cases when we multiply both X and Y individually 2 , the tens digit of 2X and 2Y can either be 2 or 3. Which means that the sum of tens digits of 2X and 2Y can be 4 (2+2) or 5 (2+3 , 3+2) or 6 (3+3).

Now on top of this just consider the maximum and minimum values the unit digit of (2X+2Y) can be which can be any even number between 0 to 16 (units of 2X or 2Y can be 0,2,4,6,8) . As long as the sum total of the unit digits is among 0,2,4,6,8 nothing needs to be added to the sum of the tens digit and it stays among 4 , 5 ,6 but when the sum total of unit digit is among 10, 12,14 ,16 then an additional 1 needs to be added to the tens digit and after adding 1 the tens digit can be 5(4+1) or 6(5+1) or 7(6+1) . So the probable values of tens digit can be 4 or 5 or 6 or 7 so a total of Four values and that's the answer
Bunuel
If the tens digit of the positive integer x and the tens digit of the positive integer y are both 6, what is the number of different possible values for the tens digit of 2x + 2y?

A. Three
B. Four
C. Five
D. Six
E. Seven

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but nowhere its given that the x and y are two digit integers. so why we are assuming that ? it could be some x=abcd6e also
Bunuel
If the tens digit of the positive integer x and the tens digit of the positive integer y are both 6, what is the number of different possible values for the tens digit of 2x + 2y?

A. Three
B. Four
C. Five
D. Six
E. Seven

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If the tens digit of the positive integer x and the tens digit of the 15 Jul 2025, 08:17
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It doesn't matter how many digits of integer they are because we are concerned about Tens digit so we would consider the digits only till tens digit from right which is the unit digit and tens digit . The tens digit when you multiply 261 with 2 and the tens digit when you multiply 234561 with 2 , in both e the case the tens digit is going to be 2. You only have to focus on 61 from the right . I hope this helps
monarchme
but nowhere its given that the x and y are two digit integers. so why we are assuming that ? it could be some x=abcd6e also
Bunuel
If the tens digit of the positive integer x and the tens digit of the positive integer y are both 6, what is the number of different possible values for the tens digit of 2x + 2y?

A. Three
B. Four
C. Five
D. Six
E. Seven

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If the tens digit of the positive integer x and the tens digit of the 19 Jul 2025, 17:41
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Since the both nos have tens digital of 6.so when we add them it can two value 2 (when there is no carryover from unit digit) and 3 (when there is carryover from unit digit).now when this 2 at unit is multiplied by 2 it can again take 2 values (4 when there is no carryover from unit digit and 5 when thereis carryover from unit digit).same goes with 3( 6 and 7 is the value it can take).so total 4 values can be of tens digit of the given expression.
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