Question

find the remainder when

Answer 1

Answer:

The remainder is 14.

I hope this helps you!

Step-by-step explanation:

The remainder when 29ⁿ is divided by 25 is the same as the remainder when 4ⁿ is divided by 25.  This is becase 29 = 25 + 4 and when we expand 29ⁿ we get lots of multiples of 25 finally followed by 4ⁿ.

So we need the remainder when 4^(67²³) is divided by 25.

Writing down powers of 4, we get 4⁵ = 1024 is one less than a multiple of 25.  Therefore 4¹⁰ leaves a remainder of 1 when divided by 25.  This is because 4⁵ = 25k - 1  =>  4¹⁰ = (4⁵)² = (25k - 1) = (a multiple of 25) + 1.

So we want to know what remainder 67²³ leaves when divided by 10.  (Just think about the last digit!)  This is the same as for 7²³.  Now 7² = 49, so 7⁴ ends in 1.  Since 7²³ = (7⁴)⁵ × 7³, the last digit of 7²³ is the same as the last digit of 7³ and that is 3.

So the last digit of 67²³ is 3.  This means 67²³ = 10n + 3, for some n.

This means

4^(67²³) = 4^(10n + 3) = (4¹⁰)ⁿ × 4³

leaves the same remainder as 4³ when divided by 25 (remember, 4¹⁰ leaves a remainder of 1).

Since 4³ = 64, the remainder when divided by 25 is 14.

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If you're familiar with congruences, this can be expressed more briefly that way.

4⁵ = 1024 ≡ -1 ( mod 25 )  =>  4¹⁰ ≡ 1 ( mod 25 )

67²³ ≡ 7²³ = (7²)¹⁰ × 7³ = (49)¹⁰ × (49)  × 7 ≡ (-1)¹⁰ × (-1)  × 7 = -7 ≡ 3 ( mod 10 )

So

29^(67²³) ≡ 4^(67²³) ≡ 4³ = 64 ≡ 14 ( mod 25 )

Answer 2

We can find the answer by,

 

ONLY JUDGING THE ONES AND TENS DIGIT!!!

 

Do you know what is meant by 29^67^23?!

 

""IT'S 29 RAISED TO THE POWER 67^23!!!""

(IT'S NOT '29 RAISED TO 67 × 23, AND THAT'S (29^67)^23. )

Oh, it'll be of a huge number. But only judge the ones and tens digit!!!

 

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Let's find the ones digit of 67^23.

 

As ones digit of 67 is 7, the possible digits which come as the ones digit of the 'powers of 67' are given IN THE ORDER:

 

7, 9, 3, 1

 

So there are 4 numbers.

By this 4, divide the exponent in 67^23, i.e., 23, and find the remainder.

We get 3.

So find the 3rd number from the order.

 

7, 9, 3, 1

 

So 3 is the ones digit of 67^23.

Let 67^23 be 'xxxxxxxxxx......xxxxx3'!!!

∴ Suppose we can say that, 29 raised to the power 67^23, is, 29 raised to the power xxxxxxxxx......xxxxx3 (Found that ones digit is 3)!!!

ALSO WE FOUND THAT 67^23 IS ""AN ODD NUMBER""!!!

 

Now it's time to find ones digit of 29^xxxxxxx......xxxxx3.

As ones digit of 29 is 9, there are only two possible numbers which come as ones digit of the 'powers of 29', and they're given IN THE RIGHT ORDER:

9, 1

 

Only 2 numbers. So the ones digit of 29ⁿ will be 9 if 'n is odd' and ones digit of 29ⁿ will be 1 if 'n is even'.

∴ Ones digit of 29^67^23 is "9", because xxxxxxx.......xxxxx3 is odd.

Oh, the divisor is '25'! So we've to find the tens digit too!!!

 

Let me do the following.

Splitting 29^xxxxxxx......xxxxx3 as the following:

29^xxxxxxxx......xxxxx3

29^xxxxxxxx......xxxxx2 × 29     [aⁿ = aⁿ ⁻ ¹ × a] [Like 2⁵ = 2⁴ × 2]

(29^2)^(xxxxxxxx......xxxxx2 / 2) × 29  [aⁿ = (a²)^(n/2)] [Like 2⁵ = (2²)^(5/2)]

Oh, no! We need to find the tens digit of xxxxxxxx......xxxxx2!!!

Let's go back!!! ↑↑↑

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Let 67^23 be 'xxxxxxxxxx......xxxxx3'!

67^23

67^20 × 67^3

(67^4)^5 × xxxx63   [67^3 = 300763]

(xxxxxx21)^5 × xxxx63  [67^4 = 20151121]

In xxxxxx21^5, the tens digits will be ones digit of 2 (which is tens digit of xxxxxx21) × 5 (which is ones digit of the exponent 5) = 10, i.e., 0. And ones digit is always 1.

∴ xxxxx01 × xxxx63

⇒ xxxxxxxx......xxxx63  [01 × 63 = 63]

∴67^23 = xxxxxxx......xxxx63

∴ xxxxxxxx......xxxxx2 will be xxxxxxxx......xxxx62

 

And xxxxxxxx......xxxxx2 / 2 = xxxxxxxx......xxxxx1.

 

(If it's an odd no. instead of 6 in xxxxxxxx......xxxx62, so xxxxxxxx......xxxxx2 / 2 will be xxxxxxxx......xxxxx6. So the tens digit was to be found.)

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OKAY, CONTINUE!!!

 

(29^2)^(xxxxxxxx......xxxxx2 / 2) × 29

(29^2)^(xxxxxxxx......xxxxx1) × 29

(x41)^(xxxxxxxx......xxxxx1) × 29   [29^2 = 841]

xxxxxxxx......xxxx41 × 29 [Like in xxxxxx21^5]

xxxxxxxx......xxxxxx89   [Last two digits of 41 × 29 = 89]

∴ No. formed by the last two digits of 29^67^23 is ''89''.

"LET'S FIND THE ANSWER!!!"

On dividing a number by 25,

"FOLLOW THESE STEPS!!!"

1. Consider the no. formed by the last two digits.
2. If it is less than 25, then it'll be the remainder.
3. If it exceeds 25, then subtract the largest multiple of 25 under that no. from it to get the number.

In 29^67^23, the no. formed by the last two digits is '89'.

 

"IT EXCEEDS 25, 50 AND 75 BUT NOT 100!!!"

   

∴ The largest multiple of 25 under 89 is '75'.

So subtract 75 from 89 to get the answer!!!

   

89 - 75 = ""14""

   

So the answer is "14".

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Hope my answer will be understood by you.

The answer is truly on my own words and it's not from any books, any websites, or not from any other sources.

   

Began answering this since 10 am today but when I finished, it's 1 pm!!!

       

Yeah, lost my precious time for you to answer!!!!!!

             

So please mark my answer as the 'brainliest' otherwise I lose my much time and, I may cry!!! :'-(

     

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Thank you. Have a nice day. :-))

   

#adithyasajeevan