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49:02 This question can be solved using other method: a+b=m^2 a-b=n^2 Multiplying the two equations,we get: a^2-b^2=(mn)^2 a^2=b^2+(mn)^2 Thus,we can say that a,b,mn form a pythagorean triplet. The smallest pythagorean triplet is (3,4,5) Thus,a=5
Grade 7 Dont know if this is different from method 1 or 2 but it looks different in my opinion Answer is 495000 abba = 110b +1001a Let a=1 Sum of the possible b values =45 Sum of a values = 10 a=2 Sum of a value= 20 a=3, 4...9 Follows similar pattern with sum= 10a Total sum of a values = 450 Total sum of b values = 45×9 =405 The sum of all 4 digit palindromes = 110×405+1001×450 = 44550+450450 = 495000 Edit: made it more clear and added the small details
495000 by By method 2 We know that 1001a+100b is a general form of palindrome Where a is not equal to 0 So possible values of a is 1 to 9. And of b is 0 to 9 So, By using this information : b's value of each case would be 45 So by factorising the expression would be 10010a + 4950 By using sigma 10010(1+2+3+4+...+9) + 4950*9 This would give 495000 Edit : I am in class 9
Answer to problem: 495000 Class: X Method as follows: First, all palindromes having middle digits '00' are taken, i.e. 1001, 2002, 3003, ..., 9009. On summing these, we can say they form 1001(1+2+3+...+9) = 45045 Now, consider all Palindromes having middle digits '11', i.e. 1111, 2112, 3113, ..., 9119. Here, we can say that each term of this sequence has been formed by adding 110 to all the corresponding '00' sequence terms. So, the sum of this series will be 110 × 9 = 990 more than the sum of '00' series = 45045 + 990 = 46035. For all subsequent palindrome series too, i.e. '22' series, '33' series and so-on, the sum will keep increasing by a common difference of 990, with so they will form an Arithmetic Progression with first term 45045, common difference 990 and number of terms 10. Applying formula for sum of A.P., we arrive at our answer, that is mentioned at the top.
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49:02 This question can be solved using other method:
a+b=m^2
a-b=n^2
Multiplying the two equations,we get:
a^2-b^2=(mn)^2
a^2=b^2+(mn)^2
Thus,we can say that a,b,mn form a pythagorean triplet.
The smallest pythagorean triplet is (3,4,5)
Thus,a=5
495450 is ans class 9 method -1 1:29:25
Grade 7
Dont know if this is different from method 1 or 2 but it looks different in my opinion
Answer is 495000
abba = 110b +1001a
Let a=1
Sum of the possible b values =45
Sum of a values = 10
a=2
Sum of a value= 20
a=3, 4...9 Follows similar pattern with sum= 10a
Total sum of a values = 450
Total sum of b values = 45×9 =405
The sum of all 4 digit palindromes = 110×405+1001×450 = 44550+450450 = 495000
Edit: made it more clear and added the small details
495000
Class 9
Other method
35:33 sir please 🙏 bring that series i am really interested in that really 😭😭😭👍👍👍👍👍👍👍👍👍👍
Sir very enjoyable session please do more
Thank you sir now I think I can crack ioqm
In class 9 using sigma
495000;Class 10;Other Method
What method can you please explain briefly
495000 by
By method 2
We know that 1001a+100b is a general form of palindrome
Where a is not equal to 0
So possible values of a is 1 to 9.
And of b is 0 to 9
So,
By using this information
:
b's value of each case would be 45
So by factorising the expression would be 10010a + 4950
By using sigma 10010(1+2+3+4+...+9) + 4950*9
This would give 495000
Edit : I am in class 9
495000, XI, M-1
Sir bahut maja aaya thank you
Happy Learning :)
Sir this series remains live at what time nd on which day as of week ,?
Answer to problem: 495000
Class: X
Method as follows:
First, all palindromes having middle digits '00' are taken, i.e. 1001, 2002, 3003, ..., 9009. On summing these, we can say they form 1001(1+2+3+...+9) = 45045
Now, consider all Palindromes having middle digits '11', i.e. 1111, 2112, 3113, ..., 9119. Here, we can say that each term of this sequence has been formed by adding 110 to all the corresponding '00' sequence terms. So, the sum of this series will be 110 × 9 = 990 more than the sum of '00' series = 45045 + 990 = 46035.
For all subsequent palindrome series too, i.e. '22' series, '33' series and so-on, the sum will keep increasing by a common difference of 990, with so they will form an Arithmetic Progression with first term 45045, common difference 990 and number of terms 10. Applying formula for sum of A.P., we arrive at our answer, that is mentioned at the top.
495000 this question is same as amc12A 2014 15th question
Sir what is differance between this series on you tube nd that you teach in course (who purchase course )
Pls explain
YT courses donot have doubts, discussion, tests, assignments, One to One Mentorship with the teachers.
Sir in the benson question can a be = 2 and b= 2 which will give 4 and 0 as perfect squares
0 isnt a square. 0 is nothing
495000 class 7 Method 2
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495000
Class 9
Other method