- Question 6
- The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is:
- The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is?
- The quotient remainder theorem (article)
Question 6
Important Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Important Deleted for CBSE Board 2024 Exams You are here Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams You are here Important Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Important Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Important Deleted for CBSE Board 2024 Exams Important Deleted for CBSE Board 2024 Exams Transcript Misc 6 Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder. Two digit numbers are 10,11,12,13, .98,99 Finding minimum number in 10,11,12,13, .98,99 which when divided by 4 yields 1 as remainder 10/4 = 22/4 11/4 = 23/4 12/4 = 3 13/4 = 31/4 So the sequence will start from 13 Finding maximum number in 10,11,12,13, .98,99 which when divided by 4 yields 1 as remainder 99/4 = 243/4 98/4 = 242/4 97/4 = 241/4 So the sequence will end at 97 Thus, the sequence starts with 13 and ends with 97 Thus, the two digit numbers which are divisible by 4 yield 1 as remainder are 13, 17, 21, 25, 93,97. This forms an A.P. as difference of consecutive terms is constant. 13, 17, 21, 25, 93,97. First term a = 13 Common difference d = 17 13 = 4 Last term l = 97 First we calculate number of terms in this AP We know that an = a + (n 1)d where an = nth term , n = number of terms, a = first term , d = common difference Here, an = last term = ...
The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is:
Correct Answer - Option 4 : 1356 By division algorithm, the two digit positive numbers which when divided by 7 yield 2 is given as: ⇒ N = 7k + 2 If k = 2 then N = 16 If k = 3 then N = 23 So on, finally, If k = 13 then N = 93 Therefore, total number of two digit positive numbers which are divided by 7 to yield 2 is 13. This forms an arithmetic progression. \( \right]\) ∴ S' = 702 The total sum is: ⇒ T = S + S' ⇒ T = 654 + 704 ∴ T = 1356
The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is?
Step -1: Find the sum of all two digit numbers which give remainder 2 when divided by 7 Two digit numbers yielding remainder 2 when divided by 7 = 7 k + 2 where k = 2 , 3 , … 1 3 = 1 6 , 2 3 , 3 0 , … 9 3 . It is an A.P. sequence whose first term ( a 1 ) = 1 6 , common difference ( d 1 ) = 7 and last term ( l 1 ) = 9 3 . Let this sequence has n 1 terms. ∴ a n 1 = l 1 = a 1 + ( n 1 − 1 ) d 1 ⇒ 9 3 = 1 6 + ( n 1 − 1 ) 7 ⇒ 7 n 1 − 7 = 7 7 ⇒ 7 n 1 = 8 4 ⇒ n 1 = 1 2 Sum of this A.P., S n 1 = 2 n 1 ( a 1 + l 1 ) = 2 1 2 ( 1 6 + 9 3 ) = 6 × 1 0 9 = 6 5 4 Step -2: Find the sum of all two digit numbers which give remainder 5 when divided by 7. Two digit numbers yielding remainder 5 when divided by 7 = 7 k + 5 where k = 1 , 2 , 3 , … 1 3 = 1 2 , 1 9 , 2 6 , … 9 6 . It is an A.P. sequence whose first term ( a 2 ) = 1 2 , common difference ( d 2 ) = 7 and last term ( l 2 ) = 9 6 . Let this sequence has n 2 terms. ∴ a n 2 = l 2 = a 2 + ( n 2 − 1 ) d 2 ⇒ 9 6 = 1 2 + ( n 2 − 1 ) 7 ⇒ 7 n 2 − 7 = 8 4 ⇒ 7 n 2 = 9 1 ⇒ n 2 = 1 3 Sum of this A.P., S n 2 = 2 n 2 ( a 2 + l 2 ) = 2 1 3 ( 1 2 + 9 6 ) = 2 1 3 × 1 0 8 = 1 3 × 5 4 = 7 0 2 Step -3: Find the sum of all two digit numbers which give remainder 2 or 5 when divided by 7. Thus, the sum of all two digit numbers that yield remainder 2 or 5 when divided by 7 , = 6 5 4 + 7 0 2 = 1 3 5 6 Hence, the correct option is D.
The quotient remainder theorem (article)
Here's the proof. Proof of the Quotient Remainder Theorem We want to prove: Given any integer A, and a positive integer B, there exist unique integers Q and R such that: A= B * Q + R where 0 ≤ R = 0 then w and f are non-negative: A/B = w + f where 0 ≤ f = 0 and any positive integer B, there exist integers Q and R such that: A= B * Q + R where 0 ≤ R = than R1 (if not we could just switch the integer pairs around). We will show that Q1 must equal Q2 and R1 must equal R2 i.e. Q and R are unique We can set the equations equal to each other B * Q1 + R1 = B * Q2 + R2 B * (Q1 - Q2) = (R2 - R1) (Q1 - Q2) = (R2 - R1)/ B since R2 >= R1 we know that R2 - R1 is >= 0 since R2 = 0 we know that R2-R1 = 0 and < 1. The only integer in that range is 0. So (R2- R1)/B= 0 ,thus R2-R1 =0 ,thus R2 = R1 also Q1-Q2 = 0 thus Q1 = Q2 Thus we have shown that Q1 must equal Q2 and R1 must equal R2 i.e. Q and R are unique Hope this makes sense We have n mod 3 = 2 n mod 5 = 1 3 and 5 are coprime so we can use the Chinese Remainder Theorem By the Chinese Remainder Theorem: Given n mod x = a n mod y = b where x and y are coprime, we have: n = (y * (y^-1 mod x) * a + x * (x^-1 mod y) * b ) mod (x * y) n = (5 * (5^-1 mod 3) * 2 + 3 * (3^-1 mod 5) * 1 ) mod (3 * 5) n = 5 * 2 * 2 + 3 * 2 * 1 mod 15 n = 26 mod 15 n = 11 Taking divisibility of numbers by 3, numbers can be expressed in 3 forms: 3q,3q+1,3q+2 (where q is the quotient obtained on dividing by 3 and 1,2 are the remainders) Let us take Case1: n=3q n+2 ...
