find tens and unit digits of 3^33 + 33^333?

Question

A)28

B)26

C)36

D)46

E)52

please help with this question.
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kajaldaryani46 · Answer

What would be the approach to this answer? I know that the concept of cyclicity is being tested here -

3^1=3
3^2=9 so on and so forth but these are simply units digits - so how do we get the tens digits here specially when the nos are large

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nick1816 · Answer

(3^{20+13}+33^{16*20+13})  Mod 100 =     mod 100

= 3^{13} (1+11^{13})

11^{13} mod 100 = 31 mod 100

( 1+11^{13} ) mod 100 = 32 mod 100

( 3^{13} ) mod 100 = 23 mod 100

3^{13} (1+11^{13})  mod100 = (32*23) mod 100 = 36 mod 100

IanStewart  It doesn't look like a GMAT question or do i miss an easy way to solve it?

IanStewart · Answer

No, I don't think you're missing anything - this question is very many miles beyond the scope of the GMAT. I did it the way you did; if you factor out a power of 3, you can at least take advantage of the familiar pattern you find in the tens and units digits of powers of 11. But unless you know 3^10 and 7^4 offhand (so you know 3^10 ends in '49' and that 49^2 ends in '01', so 3^20 ends in '01'), proving using GMAT level math that 3^20 has a remainder of 1 when you divide by 100, and that 3^13 has a remainder of 23 when you divide by 100, are both very non-trivial things to work out. From your solution it might seem to readers that you can do those things quickly, since you did them in one line, but they're not fast calculations at all, and either one alone would be far too time-consuming to be required in any GMAT problem (let alone needing to do both of them).

Learning the values of 3^10 or 7^4 would be completely pointless for any GMAT test-taker, incidentally. You'll never see an official question where knowing those things would be helpful.

Where is the question from?

kawal27 · Answer

i just know this question is made by gmac.
need expert advice  Bunuel   chetan2u  help needed.

chetan2u · Answer

Finding last two digits is always tough and when you are dealing with two different powers as in this question, it becomes even more difficult.

I would make use of choices if I were to answer it quickly..

3^{33} + 33^{333}=3^{4*8+1}+33^{332+1} 
This will have units digit as 3+3 = 6, So A and E eliminated.

3^{33} + 33^{333}=(4-1)^{33} + (32+1)^{333}=(4-1)^{33} + (4*8-1)^{333} 
So, when we look at it, the (4-1)^{33} will leave a remainder of -1^{33} or -1, and (4*8+1)^{333} will leave a remainder of 1^{333} or 1 when divided by 4.
So When divided by 4, the overall remainder will be -1+1=0.
Divisibility rule of 4 - Last two digits are divisible by 4...

Out of remaining choices only 36 is divisible by 4

OR

I would look for pattern..

3^{33} + 33^{333}=3^{33}+(30+3)^{333} 
Now in expansion of all terms of (30+3)^{333}, all except first /last two terms will have atleast 100 in it, so they will not affect the tens and hundreds digit. So we look for  3^{33}+(3^{333}+333*30^1*3^{332}=3^{33}+3^{333}+990*(3^2)^{\frac{332}{2}}=3^{33}+3^{333}+990*(9)^{166}

Let us look for a pattern for last 2-digits when raised to continuous power of 3
3^1=3...3^2=09..3^3=27..3^4=81..3^5=43...3^6=29...3^7=87...3^8=61....3^9=83
So whenever we have 3 in the units digit, the pattern becomes 
3^1=3.....3^5=43.....3^9=83......3^13=23......3^17=63......3^21=03 or 03, 43, 83, 23, 63, 03....
Thus repetition after every 20 terms..33=20+13, so 3^13 =23. Similarly 333=320+13, so again 23.
Therefore, last two digits of  3^{33}+3^{333}+990*(9)^{166}  is 23+23+990*1, as 9^even will leave 1 as units digit.
=> 23+23+90=136 or 36.

And ofcourse method of division by 100 to get last 2-digits can be done as shown above.

find tens and unit digits of 3^33 + 33^333?

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find tens and unit digits of 3^33 + 33^333?

Prep Toolkit

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