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What is the unit digit of 99?

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What is the unit digit of 99?

Therefore, Digit at unit place of 1!+ 2!+ 3!….. +99! is 3.

What is the unit digit of 3 99?

The powers of 3 have low-order digits 3, 9, 7, 1, 3, 9, 7, 1, 3, and so on. This cycle length is 4. Since 99 = 4*24 + 3, the low-order digit of 3^99 is 7.

What is the last digit in the expansion of 1 2 3 1000?

So the last digit is 1.

How do you find the digits of 3 99?

Detailed Solution

  1. Given: 399
  2. Concept Used: Unit digit of 3n 31 unit digit 3. 32 unit digit 9. 33 unit digit 7. 34 unit digit 1. 35 unit digit 3…
  3. Calculation: ⇒ 399 = (34)24 × (33) ⇒ 399 = (1)24 × 7 = (1) × 7. Unit place = 1 × 7 = 7. ∴ The place of 399 is 7. Download Soln PDF. Share on Whatsapp.
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What is the unit digit of 2 Power 99?

So we conclude that, after 40 numbers the value will be repeated will be 6 similarly after 92 the value repeated will be 6 and so the units digit of 2^100= 6.

Which is the digit in the unit place of 2 100?

Hence the units digit of 2^100= 6.

What is the unit digit of the given number?

So unit digit of the given number is 6 Solution: Here The unit place having ” 1″ so the final number is also comes ” 1″ as a unit place Hint: The last digit of any number having “9” then power having even number then unit place comes 1 and power having odd number then unit place comes 9

What is the last digit of 7 295 divided by 3 158?

Let’s divide 295 by 4 and the remainder is 3. Thus, the last digit of 7 295 is equal to the last digit of 7 3 i.e. 3. Let’s divide 158 by 4, the remainder is 2. Hence the last digit will be 9. Therefore, unit’s digit of (7 925 X 3 158) is unit’s digit of product of digit at unit’s place of 7 925 and 3 158 = 3 * 9 = 27.

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What is the unit digit of every factorial higher than 4?

You must make clear what do you mean here. If you mean what is the digit in the unit position of the integer 1! + 2! + 3! + 4! + 5! +… + 99!, then it is easy to see that for every factorial higher than 4, the unit digit is always zero.

What is the sum of 1 + 2 + 6 + 24?

So , they’ll all have 0 in their Units place. So , number at the units place in the sum of (1! + 2! + 3! + 4!) Will remain as it is in the total sum. Hence , 1 + 2 + 6 + 24 = 33 , ‘3′ will be in the units place of the whole sum.