N = 202 x 20002 x 200000002 x 2..(15 zeros)..2 x 2 ...(31 zeros) ...2 sum of the digits in this multiplication is Individual digit sum = 4 4^5 = 1024. = 7 check from options...160 has a digit sum of 7 so the answer or else u can go by traditional way
= (202*20002) = 4040404 = 16 so going by the tresnd the answwer willl be 160
for second question check the terms i think u have committed the typo eerror
ya first ans is correct ..
and i am sorry for the type error..here is ques..
10^n+(5+(17)^1/2)^n is divisible by 2^(n+2) for what whole no value of n??
Six men A,B,C,D,E,F agree with a seventh man G to provide a sum of money among them . A,B,C,D,E,F are to subscribe rs.10 each ,and G is to pay rs 3 more than the average of the seven . what is the whole sum to be provided ?
rs 73.50 rs.74 rs 73 rs 72.50
three people A,B,C weigh themselves in particular order . First A,B,C weigh themselves individually and then AB,BC, CA and ABC together respectively . the recorded weight for the last measure is 180 kgs. The average of the 7 measures is :
320 kgs 360/7 kgs 720/7 kgs. cant be determined
A certain number of trucks were required to transport 60 tons of steel wire from factory . However, it was found that since each truck could take 0.5 tons of cargo less, another 4 trucks were needed . how many trucks were initially planned to be used ?
Can some one do this question with detailed solution.. a and b have some marbles with each of them, such that the no. of marbles with b is twice that with a. If b distributes her marbles equally among certain no of bags, she is left with 41 marbles. If both a and b pool their marbles and then distribute among same no. of bags, they are left with 10 marbles. What is the least possible no. of marbles with a and b together, given that a has at least 1000 marbles? a. 3051 b. 3081 c. 3174 d. 3297 ::
Six men A,B,C,D,E,F agree with a seventh man G to provide a sum of money among them . A,B,C,D,E,F are to subscribe rs.10 each ,and G is to pay rs 3 more than the average of the seven . what is the whole sum to be provided ?
rs 73.50 rs.74 rs 73 rs 72.50
6*10+(x/7+3)=x, where x is total sum. hence x=73.50
three people A,B,C weigh themselves in particular order . First A,B,C weigh themselves individually and then AB,BC, CA and ABC together respectively . the recorded weight for the last measure is 180 kgs. The average of the 7 measures is :
320 kgs 360/7 kgs 720/7 kgs. cant be determined
here ABC =180 means... A+B+C=180..
hence total average will be 4(A+B+C)/7= option c
A certain number of trucks were required to transport 60 tons of steel wire from factory . However, it was found that since each truck could take 0.5 tons of cargo less, another 4 trucks were needed . how many trucks were initially planned to be used ?
10 15 20 25
x*y=60 (x+4) *(y-0.5)=60
where x is no of truck, y is tons /truck..solve u will get it..
Can some one do this question with detailed solution.. a and b have some marbles with each of them, such that the no. of marbles with b is twice that with a. If b distributes her marbles equally among certain no of bags, she is left with 41 marbles. If both a and b pool their marbles and then distribute among same no. of bags, they are left with 10 marbles. What is the least possible no. of marbles with a and b together, given that a has at least 1000 marbles? a. 3051 b. 3081 c. 3174 d. 3297
Kindly re check the options. According to me none of them fit. If there is none of these, then I think thats the answer. My way of solving: Let no. of marbles with a be x, then with b will be 2x total 3x For all options make set of marbles a b opt1 1017 2034 opt2 1027 2054 opt3 1058 2116 opt4 1099 2198 Now as the second case says, after they both pool in their marbles, and divide by the same number of bags, they will be left with 10. That means in each case, 3041, 3071, 3164 and 3287 will get properly distributed. Also, when b divides her marbles, then in each case 1993, 2013, 2075 and 2157 marbles get distributed properly in the same number of bags. Therefore, the two numbers in each case should have a common factor (minimum 1) but as you can see, 3041 is a prime, so first case neglected. 3071 has factors 37,83 none of whom are common with 2013 and so on for each case. Thus, no solution matches. Another way could be through equations. N1*y + 41 = 2*x and N2*y + 10 = 3*x (first one for b, second for both. N1 and N2 are different numerators and y is the number of bags) Solving we get, y = 103 As 103 is prime, y has to be either 1 or 103. We can make cases and solve, but still we do not get any solution. Hence, please check options.
Can some one do this question with detailed solution.. a and b have some marbles with each of them, such that the no. of marbles with b is twice that with a. If b distributes her marbles equally among certain no of bags, she is left with 41 marbles. If both a and b pool their marbles and then distribute among same no. of bags, they are left with 10 marbles. What is the least possible no. of marbles with a and b together, given that a has at least 1000 marbles? a. 3051 b. 3081 c. 3174 d. 3297
Kindly re check the options. According to me none of them fit. If there is none of these, then I think thats the answer. My way of solving: Let no. of marbles with a be x, then with b will be 2x total 3x For all options make set of marbles a b opt1 1017 2034 opt2 1027 2054 opt3 1058 2116 opt4 1099 2198 Now as the second case says, after they both pool in their marbles, and divide by the same number of bags, they will be left with 10. That means in each case, 3041, 3071, 3164 and 3287 will get properly distributed. Also, when b divides her marbles, then in each case 1993, 2013, 2075 and 2157 marbles get distributed properly in the same number of bags. Therefore, the two numbers in each case should have a common factor (minimum 1) but as you can see, 3041 is a prime, so first case neglected. 3071 has factors 37,83 none of whom are common with 2013 and so on for each case. Thus, no solution matches. Another way could be through equations. N1*y + 41 = 2*x and N2*y + 10 = 3*x (first one for b, second for both. N1 and N2 are different numerators and y is the number of bags) Solving we get, y = 103 As 103 is prime, y has to be either 1 or 103. We can make cases and solve, but still we do not get any solution. Hence, please check options.
Exactly thats what i was getting 103 in the end to check with the options but none of them satisfied hence i posted this question but i dont know the answer and i donno if the options are wrong or what.. anyways thanks as soon as i get the correct answer i'll post...
50^6 mod 11 which is 36*36 *36 mod 11 => 3*3*3 => 27 ...so remainder is 5
x = /4
so possibilities for x to be integer with in
-1000 y = 1,5,9,13,17........
y = /7
x = -1,-8,-15......
when y = 142 x is -993 so x also take 142 values
so thr vil be 284 values ...
Hi. my first post at pagalguy. Though your solution matches with the answer key, I have a doubt in the solution (and the key)... If I am not mistaken, the question asks the no. of pairs (x,y) which satisfy the 4x+7y=3. I solved it in the following manner:
y=(3-4x)/7 -----(i)
The general solution of x in (i) is x=7k+6 (k is an integer) Taking both the above general solution and the constraint xThus for each of the 285 values of x, we have a value of y whose magnitude will definately be less than 1000 since we have a 7 in the denominator of equation (i). So the number of pairs are 285.
If the question says find the number of integers of both x and y, then we should simply double the number of pairs = 285*2 = 570 which is not even an option.
if you consider 1, 11, 111, 1111.. upto... 11111111, each one would leave remainders 1, 2, 3,4,5,6,7,8,0 when divided by 9...
this cycle repeats for every 9 terms.
so, square of these terms would leave a remainder of 1^2, 2^2,...... which means 1, 4,0,7,7,0,4,1,0.. (the ninth term is divisible by 9. adding these and dividing it by 9 again we get remainder 6.
so we have this continuing upto 11111....45 terms
we have 46th term rem = 1 47 rem = 4 48 rem = 0 49 rem = 7
adding up we get 12
so we have 6*5 + 12 = 42
42/9 rem = 6
finally multiplying this by original 4( which was taken out as common factor) we have 24
24/9 rem = 6 Ans = 6 Also u can use the EN concept if u are aware of that
if you consider 1, 11, 111, 1111.. upto... 11111111, each one would leave remainders 1, 2, 3,4,5,6,7,8,0 when divided by 9...
this cycle repeats for every 9 terms.
so, square of these terms would leave a remainder of 1^2, 2^2,...... which means 1, 4,0,7,7,0,4,1,0.. (the ninth term is divisible by 9. adding these and dividing it by 9 again we get remainder 6.
so we have this continuing upto 11111....45 terms
we have 46th term rem = 1 47 rem = 4 48 rem = 0 49 rem = 7
adding up we get 12
so we have 6*5 + 12 = 42
42/9 rem = 6
finally multiplying this by original 4( which was taken out as common factor) we have 24
24/9 rem = 6 Ans = 6 Also u can use the EN concept if u are aware of that
Thanx dude.... I don't know about the EN concept can you explain me how to apply it???
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