1. Find the number of zeros at the end of the product:
7 × 11 × 5 × 9 × 3 × 2 × 12
Sol:
Number of 2 & 5 pairs = 1
∴ Number of zeros = 1 Answer
2. Find the number of zeros at the end of the product:
14 × 53 × 64 × 60 × 50 × 48 × 770
Sol:
14 = 2 × 7
53 ⟶ prime
64 = 2⁶
60 = 2² × 3 × 5
50 = 2 × 5 × 5
48 = 2⁴ × 3
770 = 2 × 5 × 7 × 11
By quick observation we can see that number of 5 are less than number of 2’s.
So count only number of 5 & that will be our answer.
∴ required answer = 4 Answer
3. Find the number of zeros at the end of product:
1 × 3 × 5 × 7 × 9 × 11 × ………….. × 199
Sol:
Since each term of the given product is odd number. So there will not be any ‘2’ in the given product.
Hence number of zeros = 0 Answer
4. Find the highest power of 5 in 854!
Sol:
5. Find the highest power of 7 in 1280!
Sol:
6. Find the highest power of smallest prime number that can be contained in:
1 × 2 × 3 × 4 × 5 × 6 × 7 × …………… × 222
Sol:
Given expression is equal to 222!
7. Find the number of zeros at the end of 1000!
Sol:
8. Find the number of zeros at the end of product:
1 × 3 × 5 × 7 × 9 × …………….. × 73 × 1024
Sol:
1024 = 2¹⁰ ⟵ No any power of 5
Power of 5 will come from 1 × 3 × 5 × 7 × ………. × 73
Since power of 5 i.e. 8 is less than power of 2 which is 10
∴ Number of zeros = 8 Answer
9. Find the number of zeros at the end of the product:
1 × 2 × 3 × 4 × 5 × 6 × …………. × 73 × 1024
Sol:
Product can be written as :
73! × 2¹⁰
So power of 5 comes from 73!
Since power of 2 which is 10 is less than power of 5 which is 16
∴ Number of zeros = 10 Answer
10. Find the number of zeros at the end of:
411 × 412 × 413 × …………… × 1230
Sol:
Given expression can be written as:
\(\frac{{(1 \times 2 \times 3 \times 4 \times ………. \times 410) \times (411 \times 412 \times 413 \times …………. \times 1230)}}{{(1 \times 2 \times 3 \times 4 \times ………… \times 410)}}\)
= \(\frac{{1230!}}{{410!}}\)
Now calculate number of zeros in 1230!