Hey...thakns ......for dat...
but suppose if v wann apply for ...euler's method ....den.......
hw to do dat???
plz solve this question....it came in AIMCAT-1010...
Q. The decimal number 2138 corresponds to 4132 in the number system to the base n. What is the decimal equivalent of (235)n ?
Ps:I have time's solution with me but iam not getting it.....plz give simple solution to this problem.
Thanx
plz solve this question....it came in AIMCAT-1010...
Q. The decimal number 2138 corresponds to 4132 in the number system to the base n. What is the decimal equivalent of (235)n ?
Ps:I have time's solution with me but iam not getting it.....plz give simple solution to this problem.
Thanx
2138 is lesser than 4132 .. this means base
2138/9 , remainder is 5 --> always first remainder bears last digit of base equivalent.. so ruled out ..
2138/8 , remainder is 2..
so base is 8
(235)8 = ( 157 ) 10
hope this helps
Q. The decimal number 2138 corresponds to 4132 in the number system to the base n. What is the decimal equivalent of (235)n ?
(2138 )10 => (4132)n
2*1000 + 1*100 + 3*10 + 8 = 4*n^3 + 1*n^2 + 3*n + 2*1
(2000 - 4*n^3) + (100 - n^2) + (30 - 3n) + (8-2) = 0
Only, n = 8 satisfies.
Hence (235) 8 => 157
Q. Consider the set Sn = {n,n+2,n+4,n+6,n+8} where n is a natural number between 101 to 200 (both inclusive). How many of the 100 possible sets contain multiple of 7?
thanx 
Q. Consider the set Sn = {n,n+2,n+4,n+6,n+8} where n is a natural number between 101 to 200 (both inclusive). How many of the 100 possible sets contain multiple of 7?
thanx :biggrin:
Is the Answer 70 times
Q. Consider the set Sn = {n,n+2,n+4,n+6,n+8} where n is a natural number between 101 to 200 (both inclusive). How many of the 100 possible sets contain multiple of 7?
thanx :biggrin:
check for n = 1,2,3,4,5,6,7
1 - yes , 2 - no , 3 - yes , 4 - no , 5 - yes , 6 - yes , 7 - yes
so we get (13*5) + 4 + 2 = 71
Q. Consider the set Sn = {n,n+2,n+4,n+6,n+8} where n is a natural number between 101 to 200 (both inclusive). How many of the 100 possible sets contain multiple of 7?
thanx :biggrin:
all the nos. of the form 7n, 7n+1,7n +6,7n+3,7n+5 will be a multiple of 7
hence total 7n--> 14
7n+1--> 14
7n+6-->14
7n+3-->15
7n+5--> 14
total = 71
Q1) Arun,bikas and chetakar have a total of 80 coins among them.arun triples the number of coins with the others by giving them some coins from his own collection.next,bikas repeats the same process.after this bikas now has 20 coins.find the number of coins he had the beginning?
a) 11 b) 10 c) 9 d) 12
Q2) How many triangular numbers starting with 1 and less than 1000 have the property that they are the difference of squares of two consecutive natural numbers??
a) 20 b) 21 c) 22 d) 23
Thanks..
check for n = 1,2,3,4,5,6,7
1 - yes , 2 - no , 3 - yes , 4 - no , 5 - yes , 6 - yes , 7 - yes
so we get (13*5) + 4 + 2 = 71
boss can you explain how do get (13*5) +4 +2=71
anighosh Saysboss can you explain how do get (13*5) +4 +2=71
13 * 7 = 91 .. of which only 1,3,5,6,7 satisify the condition
101 - 105 leaves remainder 3,,4,5,6,7 -- > so 4 in total
197 - 200 leaves remainder 1,2,3,4 -- > so 2 in total
13*5 + 4 + 2 = 71:cheers:
Q. Ramu started adding the digits of the page numbers of a book starting from the first page. After he finished his addition, he had obtained a sum of 1023. He missed adding the digits of one page number.
What could be the maximum number of pages in the book?
options :121, 125, 123, 128, 127
Q. Ramu started adding the digits of the page numbers of a book starting from the first page. After he finished his addition, he had obtained a sum of 1023. He missed adding the digits of one page number.
What could be the maximum number of pages in the book?
options :121, 125, 123, 128, 127
sum 1 -120 is 1023
so we can get answer as 121
but how did you get the sum as 1023. 1,2,.......9 comes out as 45. so sum of 1-99 is 450....
Aruna Kumari Saysbut how did you get the sum as 1023. 1,2,.......9 comes out as 45. so sum of 1-99 is 450....
Sum of digits from 1 to 9 = 45
Sum of all units place from 1 to 99 = 10 x45 = 450
Sum of tens digit from 1 to 99 = (1+1+..10 times) + (2+2+..10 times) + ... (9+9+9.10 times) = 10 x 45 = 450
Sum of all digits from 1 to 99 = 450 + 450 = 900
Sum of digits from 100 to 109 = 45 + 10 = 55
Sum of digits from 110 to 119 = 45+10+10= 65
Total comes from 1 to 119 = 1020
For 1 to 120 = 1023..
hope its clear .. looks lengthy .. but easy if u observe the patterns
😃 hey thank you...its clear now
Expand a^n - b^n
2138 is lesser than 4132 .. this means base
2138/9 , remainder is 5 --> always first remainder bears last digit of base equivalent.. so ruled out ..
2138/8 , remainder is 2..
so base is 8
(235)8 = ( 157 ) 10
hope this helps
Doubt: If base is 8 then digits will be 0 to 7 not '8', then how do u get 2138 in base 8 system?
tac007 SaysDoubt: If base is 8 then digits will be 0 to 7 not '8', then how do u get 2138 in base 8 system?
He must've meant that he divided 2138 by 8 to check whether he would get the last digit of 4132 or not, which he did. So his method is perfectly acceptable.
Aruna Kumari SaysExpand a^n - b^n
a^n - b^n = (a-b) * {a^(n-1) + *(b) + *(b^2) + ........ + b^(n-1) }
I hope this is what you meant by expand..........
tac007 SaysDoubt: If base is 8 then digits will be 0 to 7 not '8', then how do u get 2138 in base 8 system?
2138 is given in base 10 and 4132 is given equivalent of 2138 in another base.........Naga's method is just perfect................
