Miscellaneous - 2 | Algebra - Number System
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First 100 natural numbers in decimal system are converted to base 6 and their product if found. Find the no. of trailing 0’s in the product when converted in base 6?
Answer: 48
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Explanation :
Product of first 100 natural numbers = 100!.
When we do the same exercise in decimal system, the number of zeros is same as the highest power of 10.
Since, we are doing this exercise in base 6, the number of zeros will be same as highest power of 6.
So we have to find the highest power of 6 in 100!.
6 = 2 × 3.
∴ Highest power of 6 will be same as highest power of 3 in 100!.
Highest power of 3 is sum of all the quotients when 100 is successively divided by 2 i.e., 33 + 11 + 3 + 1 = 48.
Hence, 48.
Workspace:
If f(x, y, z) = xy + yz + zx, then f((110)2, (121)3, (1B)12) = ?
Answer: 602
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Explanation :
Converting given numbers in decimal system
(110)2 = 22 + 2 = 6
(121)3 = 32 + 2 × 3 + 1 = 16
(1B)12 = 12 + 11 = 23
∴ f(6, 16, 23) = 6 × 16 + 16 × 23 + 23 × 6 = 96 + 368 + 138 = 602.
Hence, 602.
Workspace:
A teacher wrote all the numbers from 1 to 31 in binary form. Find total number of 1’s that are there on the board?
Answer: 80
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Explanation :
(1)10 = (1)2
(31)10 = (11111)2
∴ we need to calculate number of 1’s used in all the numbers in base 2 up till 11111.
Single-digit numbers: 1
2-digit numbers: 2 + 1 = 3
3-digit numbers: 4 + 2 + 2 = 8
4-digit numbers: 8 + 4 + 4 + 4 = 20.
5-digit numbers: 16 + 8 + 8 + 8 + 8 = 48.
∴ Total number of 1’s used = 1 + 3 + 8 + 20 + 48 = 80.
Hence, 80.
Workspace:
A four-digit number N1 is written in base 11. A new four-digit number N2 is formed by rearranging the digits of N1 in any order. The difference between N1 and N2 is always divisible by?
Answer: 10
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Explanation :
Let the numbers be (abcd)11 and (dcba)11.
∴ (abcd)11 - (dcba)11 = a(113 - 1) + b(112 - 1) - c(112 - 1) - d(113 - 1)
Each of these terms is always divisible by 10 hence, the number is always divisible by 10.
Hence, 10.
Workspace:
A trader wants to measure all weights (integral) from 1 to 220 kg. using a simple balance in which weights are placed in one of the pans. What is the minimum number of weights needed?
Answer: 8
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Explanation :
We know any number can be written as sum of powers of 2.
Let the minimum weights required by n.
∴ 20 + 21 + 22 + … + 2(n-1) ≥ 220.
⇒ 2n – 1 ≥ 220
⇒ 2n ≥ 221
∴ least value of n = 8.
Hence, 8.
Workspace:
A trader wants measure (all integral weights) from 1 to 121 kg using a common balance where weights can be kept in both the pans. What is the minimum number of weights needed?
Answer: 5
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Explanation :
We know any number can be written as sum or different of powers of 3.
Let the minimum weights required by n.
∴ 30 + 31 + 32 + … + 3(n-1) ≥ 121.
⇒ (3n – 1)/2 ≥ 121
⇒ (3n – 1) ≥ 242
⇒ 3n ≥ 243
∴ least value of n = 5.
Hence, 5.
Workspace:
Find the number of 1’s in the binary notation of 2100 – 1.
Answer: 100
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Explanation :
Highest n-digit number in binary system can be written as 2n – 1 in decimal system.
∴ (2n – 1)10 = (1111….n times)2
⇒ 2100 – 1 in decimal system will have 100 1’s.
Hence, 100.
Workspace:
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