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Didn't find a shorter way: 6^17= (6*6*6*6*6)^5*6=7286^3*6 We need only the tenth digit therefore- 86*86*86*6= 7396*86 96*86=...56--- 56*6= ...36 Correct answer is B - 6

Just like cyclicity of the last digit, we can observe the cyclicity of the last 2 digits in this case : (which is possible because there is no cyclicity in the unit's digit, it is always 6) 6^2 = 36 6^3 = 16 6^4 = 96 6^5 = 76 6^6 = 56 6^7 = 36 .. and then it repeats

So for 6^17, it will have the same tens digit as 6^12, 6^7, 6^2 ... or 3

So the logic here is simple. Consider the number 6^x, lets say that you know the tens digit of this number, can you find out the tens digit of 6^(x+1) ?

What we know is that the last digit of 6^x will always be 6 (which is easy enough to see). Now the fact of the matter is that the ten's digit of 6^(x+1) is only dependent on the tens digit of 6^x.

Because Ten's digit of 6^(x+1) = 6*(Ten's digit of 6^x) + 3 (carried over during the multiplication of the units digits 6 with the new 6).

Like 6^3 = 216 So 6^4, units digit is 6 and ten's digit is 6*1+3 = 9

Hence, as soon as the ten's digit of 6^x becomes the same as the ten's digit of 6^y, the pattern of tens digit will start to repeat itself

The pattern here is 3,1,9,7,5,3,1,9,7,5,3,1,9,7,5,3,1,9,7,5,3..... The cyclicity of the pattern is five, so every 5th element in this series will be the same hence 2nd,7th,12th,17th have to be the same
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What is the tens digit of 6^17? (A) 1 (B) 3 (C) 5 (D) 7 (E) 9

There are several ways to deal with this problems some easier some harder, but almost all of them are based on the pattern recognition.

The tens digit of 6 in integer power starting from 2 (6^1 has no tens digit) repeats in pattern of 5: {3, 1, 9, 7, 5}: The tens digit of 6^2=36 is 3; The tens digit of 6^3=216 is 1; The tens digit of 6^4=...96 is 9 (how to calculate: multiply 16 by 6 to get ...96 as the last two digits); The tens digit of 6^5=...76 is 7 (how to calculate: multiply 96 by 6 to get ...76 as the last two digit); The tens digit of 6^6=...56 is 5 (how to calculate: multiply 76 by 6 to get ...56 as the last two digits); The tens digit of 6^7=...36 is 3 again (how to calculate: multiply 56 by 6 to get ...36 as the last two digits).

Hence, 6^2, 6^7, 6^12, 6^17, 6^22, ... will have the same tens digit of 3.

well, this question demands calculation to see a pattern of tens digits keep calculating till it's confirmed that u have hit a pattern. 6^1 = 6 6^2 = 36 6^3 = 216 now don't multiply 216 by 6, rather we are interested in only first two digits to know the outcome so 6^4 = 96 ( 16 x 6) 6^5 = 576 ( 96 x 6) 6^ 6 = 456 ( 76 x 6) 7^ 6 =336 ( 56 x 6) so now we have the pattern in tens digit i.e. 3 in (6^2), 1 in (6^3), 9 in (6^4), 7 in (6^5), 5 in (6^6), 3 in (6^7),

so the tens digit is 3 for the 2,7,12 and 17 times.. IMO B
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Note that when you multiply, you don't have to finish it all the way, knowing the tens digit should suffice.... Also, using the table we have we can calculate \(6^{10}\) and \(6^{17}\). We work with what we already have above/

bunuel would you please post me a link on the topic of exponents and powers from gmat math book if it has been finished..i want to learn and master way u hav solved the problem

bunuel would you please post me a link on the topic of exponents and powers from gmat math book if it has been finished..i want to learn and master way u hav solved the problem

What is the tens digit of 6^17? (A) 1 (B) 3 (C) 5 (D) 7 (E) 9

There are several ways to deal with this problems some easier some harder, but almost all of them are based on the pattern recognition.

The tens digit of 6 in integer power starting from 2 (6^1 has no tens digit) repeats in pattern of 5: {3, 1, 9, 7, 5}: The tens digit of 6^2=36 is 3; The tens digit of 6^3=216 is 1; The tens digit of 6^4=...96 is 9 (how to calculate: multiply 16 by 6 to get ...96 as the last two digits); The tens digit of 6^5=...76 is 7 (how to calculate: multiply 96 by 6 to get ...76 as the last two digit); The tens digit of 6^6=...56 is 5 (how to calculate: multiply 76 by 6 to get ...56 as the last two digits); The tens digit of 6^7=...36 is 3 again (how to calculate: multiply 56 by 6 to get ...36 as the last two digits).

Hence, 6^2, 6^7, 6^12, 6^17, 6^22, ... will have the same tens digit of 3.

Answer: B.

i have noticed that every number has 6 as the unit digit..is it the same for other numbers that they repeat each of the unit's digit throughout when it is being raised to powers of consecutive integers

What is the tens digit of 6^17? (A) 1 (B) 3 (C) 5 (D) 7 (E) 9

There are several ways to deal with this problems some easier some harder, but almost all of them are based on the pattern recognition.

The tens digit of 6 in integer power starting from 2 (6^1 has no tens digit) repeats in pattern of 5: {3, 1, 9, 7, 5}: The tens digit of 6^2=36 is 3; The tens digit of 6^3=216 is 1; The tens digit of 6^4=...96 is 9 (how to calculate: multiply 16 by 6 to get ...96 as the last two digits); The tens digit of 6^5=...76 is 7 (how to calculate: multiply 96 by 6 to get ...76 as the last two digit); The tens digit of 6^6=...56 is 5 (how to calculate: multiply 76 by 6 to get ...56 as the last two digits); The tens digit of 6^7=...36 is 3 again (how to calculate: multiply 56 by 6 to get ...36 as the last two digits).

Hence, 6^2, 6^7, 6^12, 6^17, 6^22, ... will have the same tens digit of 3.

Answer: B.

i have noticed that every number has 6 as the unit digit..is it the same for other numbers that they repeat each of the unit's digit throughout when it is being raised to powers of consecutive integers

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No. You could test that very easily yourself. Is the units digit of 2^2 equal 2? No, its 4.

• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base. • Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.

What is the tens digit of 6^17? (A) 1 (B) 3 (C) 5 (D) 7 (E) 9

There are several ways to deal with this problems some easier some harder, but almost all of them are based on the pattern recognition.

The tens digit of 6 in integer power starting from 2 (6^1 has no tens digit) repeats in pattern of 5: {3, 1, 9, 7, 5}: The tens digit of 6^2=36 is 3; The tens digit of 6^3=216 is 1; The tens digit of 6^4=...96 is 9 (how to calculate: multiply 16 by 6 to get ...96 as the last two digits); The tens digit of 6^5=...76 is 7 (how to calculate: multiply 96 by 6 to get ...76 as the last two digit); The tens digit of 6^6=...56 is 5 (how to calculate: multiply 76 by 6 to get ...56 as the last two digits); The tens digit of 6^7=...36 is 3 again (how to calculate: multiply 56 by 6 to get ...36 as the last two digits).

Hence, 6^2, 6^7, 6^12, 6^17, 6^22, ... will have the same tens digit of 3.

Answer: B.

.

Hi Bunuel,

Can we do this in following way

6^17 = (2*3)^17

(2^16*3^16) *2*3

now 2 repeats in pattern 2,4,8,6 and 3 repeats in pattern 3,9,7,1 so when we multiply 2^16*3^16 last digit is 6 now multiply 6 *6 so its 36 so tens digit is 3.

If Question like this appears in GMAT. and we are asked to find the last two digits this could be used

Rule Express Even numbers in the form ( 2^10) ^even which will have last two digits as 76 or ( 2^10) ^odd where last two digits is 24

For Odd numbers Express them 3^4k, 7 ^ 4k, 9 ^2k

Question was 6^17

So (2^ 17 ) ( 3^17)

= {(2^10)^1 * 2^7} { (3^4)^3 * 3^5}

So last two digits (2^10)^1= 24 So last two digits 2^7= 28

Last two digits of this number (3^4)^3 = (81)^3= last two digits are 41 (1 will be the last digit and second last digit will be 4 that i got by multiplying 8 *3 last two digits of this number 3^5 = 43

So (24 * 28) * (41* 43)

Don't do complete multiplication just do it till you get two digits