The sum of three consecutive integers is equal to their product. How many such possibilities are there?
UPSC-2022
Only one
Only two
Only three
No such possibility is there
We need to find sets of three consecutive integers whose sum is equal to their product. We can represent the three consecutive integers algebraically and form an equation based on the given condition. Then, we solve the equation to find the possible values for the integers.
Let the three consecutive integers be \(n-1\), \(n\), and \(n+1\).
According to the problem statement, the sum of these integers is equal to their product.
Sum: \((n-1) + n + (n+1) = 3n\)
Product: \((n-1) \times n \times (n+1) = n(n^2 - 1)\)
Equate the sum and the product:
\[ 3n = n(n^2 - 1) \]Rearrange the equation to solve for \(n\):
\[ n(n^2 - 1) - 3n = 0 \]\[ n(n^2 - 1 - 3) = 0 \]\[ n(n^2 - 4) = 0 \]Factor the expression further:
\[ n(n-2)(n+2) = 0 \]This equation gives three possible values for \(n\):
Now, find the sets of consecutive integers for each value of \(n\):
If \(n = 0\), the integers are \((0-1), 0, (0+1)\), which are \(-1, 0, 1\).
Check: Sum \((-1) + 0 + 1 = 0\). Product \((-1) \times 0 \times 1 = 0\). (Sum = Product)
If \(n = 2\), the integers are \((2-1), 2, (2+1)\), which are \(1, 2, 3\).
Check: Sum \(1 + 2 + 3 = 6\). Product \(1 \times 2 \times 3 = 6\). (Sum = Product)
If \(n = -2\), the integers are \((-2-1), -2, (-2+1)\), which are \(-3, -2, -1\).
Check: Sum \((-3) + (-2) + (-1) = -6\). Product \((-3) \times (-2) \times (-1) = -6\). (Sum = Product)
We found three distinct sets of consecutive integers that satisfy the given condition.
C) Only three. As derived, there are three sets of consecutive integers \((-1, 0, 1)\), \((1, 2, 3)\), and \((-3, -2, -1)\) that satisfy the condition.
The average weight of A, B, C is 40 kg. the average weight of B, D, E in 42 kg and the weight of F is equal to that of B What is the average weight of A, B, C, D. E and F?
UPSC-2022
40.5 kg
40.8 kg
41 kg
Cannot be determined as data is
To find the average weight of A, B, C, D, E, and F, we need to calculate the total weight of all six individuals and then divide by 6. We will use the given average weights to find the sum of weights for different groups and then combine them.
Calculate the sum of weights for A, B, C:
Given that the average weight of A, B, C is 40 kg.
So, the sum of their weights is:
\[ A + B + C = 40 \times 3 = 120 \text{ kg} \]Calculate the sum of weights for B, D, E:
Given that the average weight of B, D, E is 42 kg.
So, the sum of their weights is:
\[ B + D + E = 42 \times 3 = 126 \text{ kg} \]Use the information about F:
Given that the weight of F is equal to that of B. So, \(F = B\).
Find the total weight of A, B, C, D, E, and F:
We need to find \(A + B + C + D + E + F\). We can substitute \(F\) with \(B\).
Rearranging the terms to use the sums we already calculated:
\[ \text{Total Weight} = (A + B + C) + (D + E + B) \]Substitute the values from Step 1 and Step 2:
\[ \text{Total Weight} = 120 \text{ kg} + 126 \text{ kg} = 246 \text{ kg} \]Calculate the average weight:
The average weight of A, B, C, D, E, and F is the total weight divided by the number of individuals (which is 6).
Correct Option: C) 41 kg
Let A, B and C represent distinct non zero digits. Suppose x is the sum of all possible 3-digit numbers formed by A, B and C without repetition.
Consider the following statements:
1. The 4-digit least value of x is 1332.
2. The 3-digit greatest value of x in 888.
Which of the above statements is/are correct?
UPSC-2022
The problem asks us to analyze statements about the sum of all possible 3-digit numbers formed by three distinct non-zero digits A, B, and C without repetition. We need to determine the correct statement(s) regarding the least and greatest possible values of this sum, denoted as \(x\).
Forming 3-digit numbers: Given three distinct non-zero digits A, B, and C, the possible 3-digit numbers formed without repetition are:
Calculating the sum \(x\): The sum \(x\) of these numbers is:
\[\begin{align*} x &= (100A + 10B + C) + (100A + 10C + B) + (100B + 10A + C) \ & \quad + (100B + 10C + A) + (100C + 10A + B) + (100C + 10B + A) \end{align*}\]Collecting terms for A, B, and C:
So, \(x = 222A + 222B + 222C = 222(A + B + C)\).
Analyzing Statement 1: The 4-digit least value of \(x\) is 1332.
To find the least value of \(x\), we need to find the least possible sum of three distinct non-zero digits \(A+B+C\). Since the digits must be distinct and non-zero, the smallest possible digits are 1, 2, and 3.
So, the minimum value of \(A+B+C = 1+2+3 = 6\).
The least value of \(x\) is \(222 \times (1+2+3) = 222 \times 6 = 1332\).
This is a 4-digit number. Thus, Statement 1 is correct.
Analyzing Statement 2: The 3-digit greatest value of \(x\) is 888.
To find the greatest value of \(x\), we need to find the greatest possible sum of three distinct non-zero digits \(A+B+C\). The largest possible distinct non-zero digits are 7, 8, and 9.
So, the maximum value of \(A+B+C = 7+8+9 = 24\).
The greatest value of \(x\) is \(222 \times (7+8+9) = 222 \times 24 = 5328\).
This is a 4-digit number. The statement claims the greatest value of \(x\) is 888, which is a 3-digit number. This is incorrect. The greatest value of \(x\) is 5328.
A) 1 only. Statement 1 is correct, and Statement 2 is incorrect.
A person X wants to distribute some pens among six children A, B, C, D, E and F. Suppose A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E and six times that of F. What is the minimum number of pens X should buy so that the number of pens each one gets is an even number?
UPSC-2022
The problem asks for the minimum number of pens X should buy such that each child receives an even number of pens, given specific ratios of pens received by children A, B, C, D, E, and F. We need to find the least common multiple (LCM) of the coefficients involved in the ratios and then ensure that each child's share is an even number.
Let the number of pens received by children A, B, C, D, E, and F be \(N_A, N_B, N_C, N_D, N_E, N_F\) respectively.
From the problem statement, we have the following relationships:
\(N_A = 2 N_B \implies N_B = \frac{N_A}{2}\)
\(N_A = 3 N_C \implies N_C = \frac{N_A}{3}\)
\(N_A = 4 N_D \implies N_D = \frac{N_A}{4}\)
\(N_A = 5 N_E \implies N_E = \frac{N_A}{5}\)
\(N_A = 6 N_F \implies N_F = \frac{N_A}{6}\)
For \(N_B, N_C, N_D, N_E, N_F\) to be integers, \(N_A\) must be a multiple of 2, 3, 4, 5, and 6. Therefore, \(N_A\) must be a multiple of the LCM of (2, 3, 4, 5, 6).
Calculate the LCM of (2, 3, 4, 5, 6):
Prime factorization: \(2 = 2\), \(3 = 3\), \(4 = 2^2\), \(5 = 5\), \(6 = 2 \times 3\)
LCM \((2, 3, 4, 5, 6) = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60\)
So, \(N_A\) must be a multiple of 60. Let \(N_A = 60k\) for some positive integer \(k\).
Now, we need to ensure that the number of pens each child gets is an even number. Let's express each child's pens in terms of \(N_A\):
\(N_A\)
\(N_B = \frac{N_A}{2}\)
\(N_C = \frac{N_A}{3}\)
\(N_D = \frac{N_A}{4}\)
\(N_E = \frac{N_A}{5}\)
\(N_F = \frac{N_A}{6}\)
Substitute \(N_A = 60k\):
\(N_A = 60k\)
\(N_B = \frac{60k}{2} = 30k\)
\(N_C = \frac{60k}{3} = 20k\)
\(N_D = \frac{60k}{4} = 15k\)
\(N_E = \frac{60k}{5} = 12k\)
\(N_F = \frac{60k}{6} = 10k\)
For each of these values to be an even number, \(k\) must be chosen appropriately. All terms except \(N_D = 15k\) are already even if \(k\) is an integer. For \(N_D = 15k\) to be an even number, \(k\) must be an even number. The smallest positive even integer for \(k\) is 2.
Set \(k=2\). Then the number of pens for each child will be:
\(N_A = 60 \times 2 = 120\)
\(N_B = 30 \times 2 = 60\)
\(N_C = 20 \times 2 = 40\)
\(N_D = 15 \times 2 = 30\)
\(N_E = 12 \times 2 = 24\)
\(N_F = 10 \times 2 = 20\)
All these numbers are even.
The total minimum number of pens X should buy is the sum of pens for all children:
Total pens \( = N_A + N_B + N_C + N_D + N_E + N_F \)
Total pens \( = 120 + 60 + 40 + 30 + 24 + 20 = 294 \)
C) 294
A) 147 — This value is obtained if \(k=1\), which would make \(N_D = 15\) (odd), violating the condition that each child gets an even number of pens. The sum for \(k=1\) would be \(60+30+20+15+12+10 = 147\).
B) 150 — This value does not fit the derived ratios and even number condition.
D) 300 — This value is not the minimum. It might be obtained if \(N_A\) was a multiple of 100 or some other higher value, but it's not the smallest sum satisfying all conditions.
A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of the coins and 2 more to D. The number of coins D has now, is the smallest two digit number. How many coins does A have in the beginning?
UPSC-2022
This problem involves working backward from the final known quantity to find the initial quantity. We are given the number of coins D has and the rules by which coins were transferred from A to B, B to C, and C to D. We will reverse each transaction step by step.
D) 52
How many numbers are there between 99 and 1000 such that the digit 8 occupies the units place?
UPSC-2017
The problem asks us to find the count of numbers between 99 and 1000 (exclusive of 99 and 1000) that have the digit 8 in the units place. These numbers can be 3-digit numbers or 4-digit numbers. However, since the upper limit is 1000 (exclusive), we are only looking for 3-digit numbers.
C) 90 is the correct answer. There are 90 such numbers between 99 and 1000 where the units digit is 8.
