@techgeek2050 said:ya ryt.. sry... it should be 420 , 210 and 3.
is case me HCF will be 3...
it shud be 1...........
it shud be 1...........
@Dexian said:is case me HCF will be 3...it shud be 1...........
and i just edited it to 1 and then realised they are all greater than 1. :P
@amresh_maverick said:Sumit committed a mistake in finding the LCM of three distinct positive integers greater than 1 namely A, B and C, and found it to be 840, which is a common multiple of A, B and C all, but is not the lowest. The HCF of A, B and C is 1. Find the maximum possible value of A + B + C.
Edit: read it as minimum.. max will be 8*7+3+5=64
@abhishek.2011 said:840 is 2^3*3*5*7as lcm is less than 840 so a= 4*3*5=60b=3*5*7 =105c=4*5*7 = 140so 305 my take
hcf shd be 1
@abhishek.2011 said:840 is 2^3*3*5*7as lcm is less than 840 so a= 4*3*5=60b=3*5*7 =105c=4*5*7 = 140so 305 my take
isme hcf 5 ho gya...
sorry 4 being a JA**
sorry 4 being a JA**
@abhishek.2011 said:840 is 2^3*3*5*7as lcm is less than 840 so a= 4*3*5=60b=3*5*7 =105c=4*5*7 = 140so 305 my take
Bhai in your case HCF is not 1. Condition refuted
@amresh_maverick said:Sumit committed a mistake in finding the LCM of three distinct positive integers greater than 1 namely A, B and C, and found it to be 840, which is a common multiple of A, B and C all, but is not the lowest. The HCF of A, B and C is 1. Find the maximum possible value of A + B + C.
one of them should be 420, other should be 1, the third one is 140??
@amresh_maverick said:Sumit committed a mistake in finding the LCM of three distinct positive integers greater than 1 namely A, B and C, and found it to be 840, which is a common multiple of A, B and C all, but is not the lowest. The HCF of A, B and C is 1. Find the maximum possible value of A + B + C.
563 h kya??
A=420
B=140
C=3
LCM = 420.
A=420
B=140
C=3
LCM = 420.
@abhishek.2011 said:there are 14 consecutive numbers.find the number of ways of selecting 5 numbers such that only 2 are consecutive..
_ x _ x _ x _ x _
Here x are the choosen numbers (one of the 4 x's represents 2 numbers - consecutive one). Then we are left with 9 numbers which needs to be placed in the 5 gaps, but middle three should have aleast one
=> a + b + c + d + e = 9, such that b, c, d are atleast one
=> a + b' + c' + d' + e = 6
So, C(10, 4) ways. Also, as mentioned earlier one of the 4 x's represents two nos
=> Total no of ways = 4*C(10, 4) ways
@amresh_maverick said:Sumit committed a mistake in finding the LCM of three distinct positive integers greater than 1 namely A, B and C, and found it to be 840, which is a common multiple of A, B and C all, but is not the lowest. The HCF of A, B and C is 1. Find the maximum possible value of A + B + C.
420 140 3
sum is
563-- my take
@chillfactor said:_ x _ x _ x _ x _Here x are the choosen numbers (one of the 4 x's represents 2 numbers - consecutive one). Then we are left with 9 numbers which needs to be placed in the 5 gaps, but middle three should have aleast one=> a + b + c + d + e = 9, such that b, c, d are atleast one=> a + b' + c' + d' + e = 6So, C(10, 4) ways. Also, as mentioned earlier one of the 4 x's represents two nos=> Total no of ways = 4*C(10, 4) ways
yahi answer h hemant sir (Y)
@albiesriram said:
is it 7?
abcd/9 = bcd
or 1000a + 100b+ 10c + d = 9(100b + 10c+d)
125a = 100b + 10c +d
basically, 100b+10c + d should give 125*k
this can take upto 7 values... because 8*125 = 1000 and the resultant number will become 4digit.
abcd/9 = bcd
or 1000a + 100b+ 10c + d = 9(100b + 10c+d)
125a = 100b + 10c +d
basically, 100b+10c + d should give 125*k
this can take upto 7 values... because 8*125 = 1000 and the resultant number will become 4digit.
@techgeek2050 said:420 + 280 + 3 = 703 ?420 + 210 + 3
All these no's are divisible by 3 . so hcf=3 in your case and we need to find a triplet with hcf=1 .. so here goes my approach :
820 is a multiple of lcm .. so let our lcm be 420 which is the largest it can be ..and we want it to be the largest bcuz we want to maximise sum of three no's..
so one no is fixed at 420 .. if we take second no as 210 ..then we cant take a multiple of 2 or 3 or 5 or 7 in third no ..so we take our second no as 140 .. now we can afford to take our third no as 3 but we cant take 7.. so lets take 3 as third no :)
so three nos are 420,140 and 3 which sum up to 563 :)
Team BV--Pratik Gauri