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What is the remainder when 3^444 + 4^333 is divided by 5?

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What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 24 Feb 2020, 01:14
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Re: What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 24 Feb 2020, 03:17
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Remainder of \(3^{444} /5\) is equal to remainder of \(3^{4} /5\) ---> remainder is 1
Remainder of \(4^{333} /5\) is equal to remainder of \(4^{3} /5\) ---> remainder is 4

Remainder of \(3^{444} + 4^{333}\) divided by 5 is: (1+4) = 5, that is divisible by 5.
Thus, remainder of \(3^{444} + 4^{333}\) divided by 5 is 0 (zero)

FINAL ANSWER IS (A)
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Re: What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 24 Feb 2020, 03:19
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Solution



Method 1:


    • We know that when (\(3^4 = 81\)) is divided by 5 remainder is 1.
      o So, when \(3^{4n}\) is divided by 5, where n is an positive integer, remainder will be 1.
    • Also, when (\(4^1 = 4\)) is divided by 5 remainder is -1.
      o Therefore, when \(4^{p}\) is divided by 5, where p is a positive odd integer, remainder is will be -1.
    • Using above facts, we can write,
    • \({(\frac{{3^{444} + 4^{333}}}{5})}_{reminder}= ((\frac{3^4}{5} )^{111})_{remainder} + ((\frac{4^1}{5})^{333})_{remainder} = 1^{111} + (-1)^{333} = 1+ (-1) = 0\)
Thus, the correct answer is Option A.

Method 2:


    • Cyclicity of powers of \(3 = 4\), so unit digit of \(3^{444} = 1\)
    • Cyclicity of powers of \(4 = 2\), so unit digit of \(4^{333} = 4\)
    • Remainder of \(\frac{3^{444} + 4^{333}}{5 }\)= remainder of ((unit’s digit of \(3^{444}\) + unit’s digit of \(4^{333}\))/5)= \((\frac{1+4}{5})_{remainder} = 0\)
Thus, the correct answer is Option A.
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Re: What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 24 Feb 2020, 04:33
3 to the power 444 would have last digit as 1, while 4 to the power 333 would have last digit i.e unit place 4, so when we add both the unit digit of addition will be 5, so it will be divisible by 5 and the remainder will be 0

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Re: What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 24 Feb 2020, 08:04
use cyclicity rule we determine unit digit
3^444; 1 and 4^333; 4
unit sum ; 5
and when divided by 5 ; remainder would be 0
IMO A

What is the remainder when 3^444+4^333 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4
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Re: What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 24 Feb 2020, 09:36
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Quote:
What is the remainder when 3^444 + 4^333 is divided by 5?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


cyclicity of 3: {3,9,7,1} multiple of 4
units digits 3^444: 444/4 = last number of cycle = {1}

cyclicity of 4: {4,6} multiple of 2
units digits 4^333: 333/2 = first number of cycle = {4}

units digits of 3^444 + 4^333 = {1+4} = {5}
a number with units digit of 5 is div by 5, so remainder is 0

Ans (A)
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Re: What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 24 Feb 2020, 10:05
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3^4 = 81 leaves remainder 1 when divided by 5
3^444= (3^4)^111 will leave a remainder of 1 when divided by 5
(5-1)^333 will leave a remainder of (-1)^333 = -1 when divided by 5
3^444 + 4^333 will leave a remainder of 1-1=0 when divided by 5

IMO A

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Re: What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 24 Feb 2020, 10:17
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What is the remainder when \(3^444+4^333\) is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4

Cyclicity of 3 is 4 and that of 4 is 2.
\(3^{444}+4^{333}\) = \(3^{4k}+4^{2k' + 1}\) where k and k' are positive integers
\(3^{4k}\) would have 1 as unit digit and \(4^{2k' + 1}\) would have 4 as unit digit.

On adding \(3^{4k}\) and \(4^{2k' + 1}\) it would have 5 as unit digit which is divisible by 5.

Hence remainder is 0.

Answer A.
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Re: What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 24 Feb 2020, 10:21
3^444= 9^222
Last digit of 9^(even number)= 9

Similarly last digit of 4^(odd number)= 4
therefore last digit of 4^333= 4

now R(9/5) + R(4/5)= R98/5)= 3
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Re: What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 24 Feb 2020, 10:22
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What is the remainder when 3^444+4^333 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4

3 has a cycle of 4
3^1 =3
3^(2 )= 9
3^3 = 7
3^4 = 1

4 has a cycle of 2
4^1 = 4
4^2 = 6
4^3 = 4

Now,3^444+4^333
= 3^(111)*4+4^(83*4+1)
= 1(Unit digit of 3 at cycility of 4) + 4(Unit digit of 4 at cycility of 1)
= 1 + 4
= 5
Therefore, remainder of 5/5 =0
Answer: A
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Re: What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 24 Feb 2020, 12:04
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The best way to solve this problem is by decomposing the exponents from the base, that is:

3^444 = (3^4)^100 * (3^4)^10 * (3^4)^1
4^333 = (4^3)^100 * (4^3)^10 * (4^3)^1

According to this, the answer is A since the units digits of the sum is 0.
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Re: What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 24 Feb 2020, 22:34
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What is the remainder when \(3^{444} + 4^{333}\) is divided by 5?

\(3^{444} + 4^{333} = 9^{222} + 4^{333}\)

—> \(( 10–1)^{222} + (5–1)^{333} = 10m + 1 + 5n —1 = 5*(2m +1)\)

The remainder is zero

The answer is A.

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Re: What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 25 Feb 2020, 01:17
Any number that has 5 or 0 as its units digit is divisible by 5. As such, any number that has any other digit in the units place will give a remainder of 1 or 2 or 3 or 4 when divided by 5. Remember, the biggest remainder you can get when you divide any number by 5, is 4, while the smallest is 0.

The cyclicity of units digit of numbers ending with 3, is 4 – any number ending with 3 will always end with 3,9,7 or 1 depending on the exponent to which it is raised.
If the exponent is a multiple of 4, the units digit will always be 1. In our case, \(3^{444}\) will definitely end with 1.

The cyclicity of units digit of numbers ending with 4, is 2 – any number ending with 4 will always end with 4 or 6. If the power is odd, the number will have 4 as the units digit and if the power is even, the units digit will be 6.
In our case, \(4^{333}\) will have 4 as the units digit since the power is odd.

Therefore, \(3^{444}\) +\( 4^{333}\) gives us a number ending with 5 which means we have a multiple of 5. The remainder when this number is divided by 5, will be ZERO.

The correct answer option is A.

Hope that helps!
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Re: What is the remainder when 3^444 + 4^333 is divided by 5?  [#permalink]

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New post 25 Feb 2020, 02:42
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Well, people, this one can be a pain to solve. There are a lot of little steps along the way, even if you know the concept, to trip up on. Yuck.

Fundamentally, you first need to understand that when you increase the powers of the numbers 3 or 4, they begin to show a pattern when it comes to their last digit

3 to the power 1, 2, 3 ,4 becomes 3, 9, 27, 81 ....something ending with 3... something ending with 9 and the pattern continues on and on.... Endless last digit blocks of 3, 9, 7, 1

4 to the power 1, 2, 3 becomes 4, 16, 64.... something ending with 6... and the pattern continues.... Endless last digit blocks of 4, 6

Once you know this the rest is just calm calculation.

What's the last digit of 3 to the power 444 ---> 444/4 leaves you with a remainder of 0 --- meaning you are at the 4th "house" on the block. That 4th house ends in a 1.

Similarly, to find the last digit of 4 to the power 333, just divide 333/2. You don't even have to solve it. Recognize it will not evenly divide and therefore it will have to be an Odd House ending in 4.

Now that we know both the last digits, we can simply add them before dividing by 5.

And guess what? 1 + 4 divided by 5 leaves a remainder of 0!
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Re: What is the remainder when 3^444 + 4^333 is divided by 5?   [#permalink] 25 Feb 2020, 02:42
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