Explanation for the Arithmetic Quiz 04

This is the detailed explanation for the Explanation for the Arithmetic Quiz 04. Each question’s solution is broken down step-by-step to help you understand the core concepts of remainders, HCF, and LCM.

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1. Find the least number which will leaves remainder 5 when divided by 8, 12, 16 and 20.

Answer: 245

1. Find the LCM (Least Common Multiple) of the divisors. The number must be a multiple of the LCM of 8, 12, 16, and 20.
   • Prime factorization:
     • 8 = 23
     • 12 = 22×3
     • 16 = 24
     • 20 = 22×5
   • LCM = 24×3×5=16×3×5=240.
2. Add the remainder. The number must be 5 more than a multiple of 240.
   • Number = LCM + Remainder = $240+5=245$.

2. Three numbers are in ratio 1:2:3 and HCF is 12. The numbers are:

Answer: 12, 24, 36

1. Understand the relationship between numbers, HCF, and ratio. If numbers are in a given ratio, and their HCF is known, the numbers themselves are the HCF multiplied by each term in the ratio.
2. Calculate the numbers.
   • First number = $1\times 12=12$
   • Second number = $2\times 12=24$
   • Third number = $3\times 12=36$

3. Find the remainder when 6799 is divided by 7.

Answer: 6

1. Divide the number by the divisor.
   • $6799÷7$
2. Perform the division.
   • $6799=(7\times 971)+6$.
3. Identify the remainder. The remainder is 6.

4. After the division of a number successively by 3, 4 and 7, the remainder obtained is 2, 1 and 4 respectively. What will be remainder if 84 divide the same number?

Answer: 53

1. Find the smallest number that fits the conditions. This is a reverse-calculation problem.
   • Start from the last division. A number divided by 7 gives a remainder of 4. Let’s take the smallest such number: $7\times 1+4=11$.
   • This number (11) was the quotient from the previous step. A number divided by 4 gives a remainder of 1 and a quotient of 11. The number is $(4\times 11)+1=45$.
   • This number (45) was the quotient from the first step. A number divided by 3 gives a remainder of 2 and a quotient of 45. The number is $(3\times 45)+2=135+2=137$.
2. Divide the found number by 84.
   • $137÷84$
3. Find the remainder.
   • $137=(84\times 1)+53$. The remainder is 53.

5. Find the remainder when $73\times 75\times 78\times 57\times 197\times 37$ is divided by 34.

Answer: 30

1. Find the remainder for each number when divided by 34. This is modular arithmetic.
   • $73÷34⟹$ remainder is $73-(34\times 2)=73-68=5$.
   • $75÷34⟹$ remainder is $75-(34\times 2)=75-68=7$.
   • $78÷34⟹$ remainder is $78-(34\times 2)=78-68=10$.
   • $57÷34⟹$ remainder is $57-34=23$. (or -11)
   • $197÷34⟹$ remainder is $197-(34\times 5)=197-170=27$. (or -7)
   • $37÷34⟹$ remainder is $37-34=3$.
2. Multiply the remainders.
   • $5\times 7\times 10\times 23\times 27\times 3$.
3. Simplify the multiplication by taking remainders again.
   • $(5\times 7)(\mathrm{mod}34)=35(\mathrm{mod}34)⟹1$.
   • $10\times 23=230$. $230÷34=(34\times 6)+26=26$.
   • $27\times 3=81$. $81÷34=(34\times 2)+13=13$.
   • Now, we have $1\times 26\times 13=338$.
4. Find the remainder of the final product.
   • $338÷34=(34\times 9)+32=32$.
   • Wait, the answer is 30. Let’s re-calculate.
   • $73\equiv 5(\mathrm{mod}34)$
   • $75\equiv 7(\mathrm{mod}34)$
   • $78\equiv 10(\mathrm{mod}34)$
   • $57\equiv 23(\mathrm{mod}34)$
   • $197\equiv 27(\mathrm{mod}34)$
   • $37\equiv 3(\mathrm{mod}34)$
   • Product of remainders: $5\times 7\times 10\times 23\times 27\times 3(\mathrm{mod}34)$
   • $5\times 7=35\equiv 1(\mathrm{mod}34)$
   • $10\times 23=230\equiv 26(\mathrm{mod}34)$
   • $27\times 3=81\equiv 13(\mathrm{mod}34)$
   • $1\times 26\times 13(\mathrm{mod}34)=338(\mathrm{mod}34)$.
   • $338=340-2=34\times 10-2$. The remainder is 32. The provided answer of 30 is likely incorrect.

6. What is the least number of soldiers that can be drawn up in troops of 12, 15, 18 and 20 soldiers and also in form of a solid square?

Answer: 900

1. Find the LCM of the troop sizes. The number of soldiers must be a multiple of 12, 15, 18, and 20.
   • LCM of 12, 15, 18, 20.
     • 12 = 22×3
     • 15 = $3\times 5$
     • 18 = 2×32
     • 20 = 22×5
   • LCM = 22×32×5=4×9×5=180.
2. Check if the LCM is a perfect square. 180 is not a perfect square.
3. Multiply the LCM by the missing factors to make it a perfect square. To be a perfect square, all prime factors in its prime factorization must have even powers. The prime factors of 180 are 22,32,51. The factor ‘5’ has an odd power. We need to multiply by 5 to make the power even (52).
   • Number = $180\times 5=900$.
4. Confirm the number is a perfect square. 900=302.

7. Find the remainder when 496 is divided by 6.

 

Answer: 4

1. Examine the pattern of powers of 4.
   • 41=4. $4÷6⟹$ remainder is 4.
   • 42=16. $16÷6=(6\times 2)+4⟹$ remainder is 4.
   • 43=64. $64÷6=(6\times 10)+4⟹$ remainder is 4.
2. Identify the pattern. Any positive integer power of 4 (4n for $n\ge 1$) when divided by 6 will have a remainder of 4.

8. Find the remainder when 2256 is divided by 17.

Answer: 1

1. Use Fermat’s Little Theorem. The theorem states that if ‘p’ is a prime number, then for any integer ‘a’, the number ap−a is an integer multiple of ‘p’. In other words, ap≡a(modp). If ‘a’ is not divisible by ‘p’, then ap−1≡1(modp).
2. Apply the theorem to the problem.
   • Here, $a=2$ and $p=17$. 17 is a prime number, and 2 is not divisible by 17.
   • According to the theorem, 217−1=216≡1(mod17).
3. Use the property of exponents. We need to find the remainder of 2256. We can express 256 in terms of 16.
   • $256=16\times 16$.
   • 2256=2(16×16)=(216)16.
4. Find the remainder.
   • (216)16≡(1)16(mod17)≡1(mod17).
   • The remainder is 1.

9. 7n−6n, where n is an integer >0, is divisible by

Answer: All of these

1. Test the expression for n=1.
   • 71−61=7−6=1. This doesn’t help much.
2. Test for n=2.
   • 72−62=49−36=13. This is divisible by 13.
3. Test for n=3.
   • 73−63=343−216=127. This is divisible by 127.
4. Use the algebraic formula: an−bn=(a−b)(an−1+an−2b+…+bn−1).
   • 7n−6n=(7−6)(7n−1+…+6n−1)=1×(…). This shows it’s always divisible by $(7-6)=1$. Not helpful.
5. Let’s reconsider the question and options.
   • The options are 13, 127, and 559.
   • For $n=2$, the value is 13.
   • For $n=3$, the value is 127.
   • What is the divisibility of the expression in general?
   • For any odd n, an+bn is divisible by $a+b$.
   • For any even n, an−bn is divisible by $a-b$ and $a+b$.
   • The question has $a=7,b=6$. So $a-b=1$. The expression is not guaranteed to be divisible by $a+b=13$. For example, for $n=1$, 71−61=1 which is not divisible by 13.
   • There seems to be a mistake in the question’s premise or the “All of these” option. Let’s assume the question meant a specific value of n, or a different expression.
   • Let’s check the number 559. $559=13\times 43$.
   • The question is flawed. The expression 7n−6n is divisible by 13 for even n, and by 127 for n=3, and by 1 for any n. The phrase “is divisible by” implies it is always divisible. This is not true for all ‘n’. The answer “All of these” is highly suspect.

10. Let N = 1421*1423*1425. What is the remainder when N is divided by 12?

Answer: 9

1. Find the remainder of each number when divided by 12.
   • $1421=(12\times 118)+5$. Remainder is 5.
   • $1423=(12\times 118)+7$. Remainder is 7.
   • $1425=(12\times 118)+9$. Remainder is 9.
2. Multiply the remainders and find the remainder of the product.
   • $(5\times 7\times 9)(\mathrm{mod}12)$
   • $5\times 7=35$. $35÷12=(12\times 2)+11$. Remainder is 11.
   • $11\times 9=99$. $99÷12=(12\times 8)+3$. Remainder is 3.
   • Wait, the provided answer is 9. Let’s re-calculate.
   • $5\times 7\times 9=315$.
   • $315÷12$. $315=(12\times 26)+3$. Remainder is 3.
   • Let’s try another method.
   • $1421\equiv 5(\mathrm{mod}12)$
   • $1423\equiv 7(\mathrm{mod}12)$
   • $1425\equiv 9(\mathrm{mod}12)$
   • $1421\times 1423\times 1425\equiv 5\times 7\times 9(\mathrm{mod}12)$
   • $\equiv 35\times 9(\mathrm{mod}12)$
   • $\equiv 11\times 9(\mathrm{mod}12)$
   • $\equiv 99(\mathrm{mod}12)$
   • $99=12\times 8+3$. Remainder is 3.
   • The provided answer of 9 is incorrect based on the numbers given. The correct answer is 3.