Can someone please elaborate a bit on the bold part. pardon my ignorance.. :)
see the number is of the form 7x = 8y+3 = 9z+6 (that is its divisable by 7, leaves a remainder 3 and 6 when divided by 8 and 9 respectively) now find a no. of the form 8y+ 3 = 9y+6 this means the no. 9y+3 shud be divisable by 8....at y= 5 we get 48 which is divisable by 8 hence the no. is 48+3 = 51 so we can wrt it as LCM(8,9)k + 51 = 72k+51...similarly do it for 7x = 72y + 51...u can see tht at y=6 we get 483 which is divisable by 7 hence the no. is of the form LCM(7,72)k + 483 = 504k + 483...hence the remainder is 483
see the number is of the form 7x = 8y+3 = 9z+6 (that is its divisable by 7, leaves a remainder 3 and 6 when divided by 8 and 9 respectively) now find a no. of the form 8y+ 3 = 9y+6 this means the no. 9y+3 shud be divisable by 8....at y= 5 we get 48 which is divisable by 8 hence the no. is 48+3 = 51 so we can wrt it as LCM(8,9)k + 51 = 72k+51...similarly do it for 7x = 72y + 51...u can see tht at y=6 we get 483 which is divisable by 7 hence the no. is of the form LCM(7,72)k + 483 = 504k + 483...hence the remainder is 483
The lines Y=3x and X+Y = 40 and the X-axis bound a triangular area. Find the total number of points on or inside this triangle with integral coordinates..
The lines Y=3x and X+Y = 40 and the X-axis bound a triangular area. Find the total number of points on or inside this triangle with integral coordinates..
i was also getting the answer in three hundred something but the options are far away from our answer..
The lines Y=3x and X+Y = 40 and the X-axis bound a triangular area. Find the total number of points on or inside this triangle with integral coordinates..
The lines Y=3x and X+Y = 40 and the X-axis bound a triangular area. Find the total number of points on or inside this triangle with integral coordinates..
The two lines will intersect at (10,30)
Divide the triangle in two regions one from ( 0,0 ) to ( 10,30 ) and another from ( 10,30 ) t0 ( 0,40 )
Now count number of points from ( 0,0 ) to ( 10,30 ) ( 0,0 ) - 1 point (1,0) ( 1,1 ) ( 1,2 ) (1,4 ) - 4 points ( 2,0) ( 2,1) ( 2,2 ).........( 2,6 ) - 7 points . . . .till ( 10,0 ) (10,1 ) (10,2 ).......... ( 10,30 ) - 31 points
=> total number of points for R1 = 1+4+7+..31 => total 11 terms in A.P. with d =3 => number of points = 176
Similarly for region 2 number of points = 1+2+3 +.....30 = 465
The lines Y=3x and X+Y = 40 and the X-axis bound a triangular area. Find the total number of points on or inside this triangle with integral coordinates..
:banghead::banghead:
triangle will be in first quadrant. (0,0) -- (0,0) (1,0) -- (1,3) (2,0) -- (2,6) . . (10,0) -- (10,30) (11,0) -- (11,29) (12,0) -- (12,28 ) . . (40,0) -- (40,0) total points = (1+4+7+ .. + 31) + 30 +29+28+..+1= 176 + 465 = 641
Q>Two numbers have 36 factors each and HCF of these two numbers is 36. What is the minimum possible LCM of these two numbers if the power of any prime factor in these two numbers is not more than 3?
Q>Two numbers have 36 factors each and HCF of these two numbers is 36. What is the minimum possible LCM of these two numbers if the power of any prime factor in these two numbers is not more than 3?
Q> Find the digit at the tens place of the number N = 7^281 3^264.
1. 3 2. 4 3. 5 4. 6 5. 7
This can rewritten as 21^264 * 7^17... 7 raise to power repeats the digit after four powers 7^1 = 07, 7^2 = 49, 7^3 = 43, 7^4 = 01, 7^5 = 07 and so on So, 7^17 ends in 07 21 raise to power also repeats but after five powers 21^1 = 21, 21^2 = 41, 21^3 = 61, 21^4 = 81, 21^5 = 01 21^6 = 21 and so on So, 21^265 ends in 81 So 81 * 07 is 67... i.e. ten's digit is 6
Also can be solved by standard Euler's theorem concept... Ten's digit can be found by finding remainder of dividing N by 100 Euler No of 100 is 100*(1-1/2)*(1-1/5)=40 So p^40 divide by 100 will leave a remainder 1 where p is a prime number i.e.7^280:100 is 1 hence 7^1:100 is 07 and 3^240:100 is 1 and 3^24 is 729^4:100 is equal to 29^4:100 41^2:100 i.e. 81... So 81*07 i.e. 67... i.e. ten's digit is 6
x = 2^2 * 3^2 * ........... y = 2^2 * 3^2 * ...........
Now both have factors as 9 now for factors to be 36 we must have 4 from other prime numbers........
which can be in the format -> p*q or p^3 and q^3 Now for p*q both will have same p*q which will violate the condition of HCF so possible combination is...........
Q>Two numbers have 36 factors each and HCF of these two numbers is 36. What is the minimum possible LCM of these two numbers if the power of any prime factor in these two numbers is not more than 3?
36 factors i.e. 2*2*3*3 since no powers of greater than 3.... i.e. can be rewritten as 4*3*3 so prime nos has powers of 3,2 and 2 respectively... Also since 36 is hcf so both nos should have 2^2 and 3^2 i.e. two primes are 2 and 3 in each no and third prime is one will be 5 and in other 7 to make LCM the least... i.e. First no 2^3*3^2*5^2 Second no 2^2*3^3*7^2