Find the digit at the ten's place of the number N = 7281 * 3264.
(A)3
(B)4
(C)5
(D)6
(E)7
The OA is D.
Don't look at the soln. before you solve it yourself.
[spoiler]In other words the question asks to find out the remainder
when the expression is divided by 100 and then take the left
most digit as the final result.
Also, 720k when divided by 100 gives remainder 1.
(Here K is a positive integer)
720k = (49)10k = (50 - 1)10k.
In the expansion of (50 - 1)10k, every term except the last term,
i.e. 1 will be divisible by 100.
Therefore, when 720k is divided by 100, the remainder is 1.
Even, 320K when divided by 100 gives remainder 1.
So, 7281 × 3264 by 100 ⇒ 7 × 34 by 100 ⇒ 67 by 100.
Hence, the digit at the tens place is 6.[/spoiler]
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Digits at ten's place
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harsh.champ
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It takes time and effort to explain, so if my comment helped you please press Thanks button 
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shashank.ism
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Hmm.. quite a difficult one to be approached by onself but solution made to me understand a new concept..nice question.harsh.champ wrote:Find the digit at the ten's place of the number N = 7281 * 3264.
(A)3
(B)4
(C)5
(D)6
(E)7
The OA is D.
Don't look at the soln. before you solve it yourself.
[spoiler]In other words the question asks to find out the remainder
when the expression is divided by 100 and then take the left
most digit as the final result.
Also, 720k when divided by 100 gives remainder 1.
(Here K is a positive integer)
720k = (49)10k = (50 - 1)10k.
In the expansion of (50 - 1)10k, every term except the last term,
i.e. 1 will be divisible by 100.
Therefore, when 720k is divided by 100, the remainder is 1.
Even, 320K when divided by 100 gives remainder 1.
So, 7281 × 3264 by 100 ⇒ 7 × 34 by 100 ⇒ 67 by 100.
Hence, the digit at the tens place is 6.[/spoiler]
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harshavardhanc
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Harsh, can you please verify the question / answer options once again. Because the product here (7281 * 3264) is 23765184, and the tens digit is 8.
Regards,
Harsha
Harsha
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shashank.ism
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harshavardhanc, I think the question is 7^281*3^264. Try it you will get to understand the solution.Its a nice concept,\.harshavardhanc wrote:Harsh, can you please verify the question / answer options once again. Because the product here (7281 * 3264) is 23765184, and the tens digit is 8.
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shashank.ism wrote:Mr. HArsh, how did you arrive at the step 7^20k when divided by 100 gives remainder 1? Kindly explainharsh.champ wrote:Find the digit at the ten's place of the number N = 7281 * 3264.
(A)3
(B)4
(C)5
(D)6
(E)7
The OA is D.
Don't look at the soln. before you solve it yourself.
[spoiler]In other words the question asks to find out the remainder
when the expression is divided by 100 and then take the left
most digit as the final result.
Also, 720k when divided by 100 gives remainder 1.
(Here K is a positive integer)
720k = (49)10k = (50 - 1)10k.
In the expansion of (50 - 1)10k, every term except the last term,
i.e. 1 will be divisible by 100.
Therefore, when 720k is divided by 100, the remainder is 1.
Even, 320K when divided by 100 gives remainder 1.
So, 7281 × 3264 by 100 ⇒ 7 × 34 by 100 ⇒ 67 by 100.
Hence, the digit at the tens place is 6.[/spoiler]
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Matt@VeritasPrep
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paiboi782 wrote:7²�ᴷ => 49¹�ᴷ => 2401�ᴷshashank.ism wrote:Mr. HArsh, how did you arrive at the step 7^20k when divided by 100 gives remainder 1? Kindly explainharsh.champ wrote:Find the digit at the ten's place of the number N = 7281 * 3264.
(A)3
(B)4
(C)5
(D)6
(E)7
The OA is D.
Don't look at the soln. before you solve it yourself.
[spoiler]In other words the question asks to find out the remainder
when the expression is divided by 100 and then take the left
most digit as the final result.
Also, 720k when divided by 100 gives remainder 1.
(Here K is a positive integer)
720k = (49)10k = (50 - 1)10k.
In the expansion of (50 - 1)10k, every term except the last term,
i.e. 1 will be divisible by 100.
Therefore, when 720k is divided by 100, the remainder is 1.
Even, 320K when divided by 100 gives remainder 1.
So, 7281 × 3264 by 100 ⇒ 7 × 34 by 100 ⇒ 67 by 100.
Hence, the digit at the tens place is 6.[/spoiler]
2401 to any (positive integer) power is going to end in 01, so it will always have remainder 1 when divided by 100.
This is not at all relevant to the GMAT in 2017 (or any prior year).
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Matt@VeritasPrep
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A much easier solution, by the way:
7281 * 3264 will have the same tens digit as 81 * 64.
81 * 64 = (80 + 1) * (60 + 4) = 4800 + 60 + 320 + 4
The tens digit comes from 60 + 320 = 380, so our tens digit is 8.
7281 * 3264 will have the same tens digit as 81 * 64.
81 * 64 = (80 + 1) * (60 + 4) = 4800 + 60 + 320 + 4
The tens digit comes from 60 + 320 = 380, so our tens digit is 8.
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Matt@VeritasPrep
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If you're wondering how I can simplify like that, consider a related problem.
Say I have one big number ending in 81 and another big number ending in 64.
I can write the first as 100x + 81, since everything greater than 81 is in the hundreds place or further left. (For instance, 7281 = 72*100 + 81.)
In much the same way, I can write the second number as 100y + 64.
When I multiply them together, I get:
(100x + 81) * (100y + 64) =>
10000xy + 8100y + 6400x + 81*64
Everything other than 81*64 is too big to contribute to the tens digit, so the tens digit will be generated solely by 81*64.
Say I have one big number ending in 81 and another big number ending in 64.
I can write the first as 100x + 81, since everything greater than 81 is in the hundreds place or further left. (For instance, 7281 = 72*100 + 81.)
In much the same way, I can write the second number as 100y + 64.
When I multiply them together, I get:
(100x + 81) * (100y + 64) =>
10000xy + 8100y + 6400x + 81*64
Everything other than 81*64 is too big to contribute to the tens digit, so the tens digit will be generated solely by 81*64.
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Matt@VeritasPrep
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If the problem is actually 7²�¹ * 3²�� - which would be pretty sloppy posting, but fair enough, formatting errors happen - then I'd do it somewhat differently.
7²�¹ * 3²�� =>
7¹� * 7²�� * 3²�� =>
7¹� * 21²��
From here, we can work with a neat property of 21. As we cycle through its powers, we notice that their units digits follow a pattern:
21¹ = 21
21² = 441
21³ = 9261
21� = 194481
21� = 4084101
(I'll show why this is happening in a sec.)
Once we get to the 01, we know that any power of that base is going to end in 01. So we can say that
21�ᴷ = ...01, for any integer K ≥ 1.
From here, we've got
7¹� * 21� * 21²�� =>
7¹� * 21� * (21�)�² =>
7¹� * 21� * ...01 =>
7¹� * ...81 * ...01
Now we need to work with 7¹�. Since 7� has the same property as 21� (7� = 2401), we can cut out any 7�ᴷ:
7¹� * ...81 * ...01 =>
(7�)� * 7 * ...81 * ...01 =>
...01 * 7 * ...81 * ...01
The only parts of this that will generate the tens digit are the 7 and the ...81. 7 * ...81 = ...567, so our tens digit is 6.
7²�¹ * 3²�� =>
7¹� * 7²�� * 3²�� =>
7¹� * 21²��
From here, we can work with a neat property of 21. As we cycle through its powers, we notice that their units digits follow a pattern:
21¹ = 21
21² = 441
21³ = 9261
21� = 194481
21� = 4084101
(I'll show why this is happening in a sec.)
Once we get to the 01, we know that any power of that base is going to end in 01. So we can say that
21�ᴷ = ...01, for any integer K ≥ 1.
From here, we've got
7¹� * 21� * 21²�� =>
7¹� * 21� * (21�)�² =>
7¹� * 21� * ...01 =>
7¹� * ...81 * ...01
Now we need to work with 7¹�. Since 7� has the same property as 21� (7� = 2401), we can cut out any 7�ᴷ:
7¹� * ...81 * ...01 =>
(7�)� * 7 * ...81 * ...01 =>
...01 * 7 * ...81 * ...01
The only parts of this that will generate the tens digit are the 7 and the ...81. 7 * ...81 = ...567, so our tens digit is 6.
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Matt@VeritasPrep
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As a footnote, let me add the pattern with powers of 21.
Notice that 21 * 21 = (20 + 1) * 21 = 20*21 + 21 = 420 + 21 = 441. See how we've got 20 at the end of the first number?
Now move to the next power: 441 * 21 => (440 + 1) * 21 => 440*21 + 21 => 9240 + 21 => 9261.
Now let's say we have another multiple of 20 before the +21. Our multiples of 20 will always end in 00, 20, 40, 60, or 80. Any time we add 21 to that, we'll get 21, 41, 61, 81, or 01. So our powers of 21 will continue to cycle like this, and if we only need to consider the tens digit, we can use this pattern to get it.
Notice that 21 * 21 = (20 + 1) * 21 = 20*21 + 21 = 420 + 21 = 441. See how we've got 20 at the end of the first number?
Now move to the next power: 441 * 21 => (440 + 1) * 21 => 440*21 + 21 => 9240 + 21 => 9261.
Now let's say we have another multiple of 20 before the +21. Our multiples of 20 will always end in 00, 20, 40, 60, or 80. Any time we add 21 to that, we'll get 21, 41, 61, 81, or 01. So our powers of 21 will continue to cycle like this, and if we only need to consider the tens digit, we can use this pattern to get it.
