See, N = 10*(abcdef) + 5 , let abcdef = MSo, N = 10*(M) + 5Now, If M = 7k + 2 , then N = 70k + 25 => NOT divisible by 35 But, If M = 7k + 3, then N = 70k + 35 => divisible by 35 How I got it ?N mod 5 = 10*M mod 5 + 5 mod 5 = 0 N mod 7 = 10*M mod 7 + 5 mod 7 = (10 mod 7)*(M mod 7) + 5 = 3*(R) + 5 For N mod 7 = 0 , R = 3 satisfies => M mod 7 = 3 => M = 7k + 3=> abcdef = 7k + 3
Is the answer 9 ??Let the number be abcdef5Now a*10^6 + b*10^5 + c*10^4 + d*10^3 + e*10^2 + f*10 +5 mod 7 = 0We can see that 5*10^6 mod 7 = 55*10^5 mod 7 = 45*10^4 mod 7 = 65*10^3 mod 7 = 25*10^2 mod 7 = 35*10^1 mod 7 = 1(Taken 5 because we are required to use 5 in the number and without which it was getting difficult to solve further)Now digits can be 5 or 7 only.=> We have to make diff possible arrangements such that abcdef5 mod 7 = 0Now, All 7s --> not possibleone 5---> 5+2 = 7 --> 1 possibilitytwo 5---> 5+4+5=14, 5+6+3 =14 --> 2 waysthree 5---> 5+1+2+6=14, 5+1+3+5=14, 5+3+2+4=14 --> 3 waysfour 5---> 5+5+4+6+1 =21, 5+6+5+3+2 =21 --> 2 waysfive 5 ---> 5 + 1+2+3+4+6 = 21 --> 1 wayHence a total of 9 possibilities ? ..Nos are: 7775775, 5577775, 7757575, 7755755, 57777555, 7575575, 5557755, 5755575, 7555555 .. P.S. Only I know kaha kaha se dhua nikla iske baad..
4 Cars A,B,C,D Start together A&B; at higher speed both having same speed, C & D lower speed, by the time C & D cover 80 Kms, A & B reach CLOCK TOWER (Name of a place), B Then rests for an hour and C & D reach there (at Clock Tower in 1 hour the time when B Rests) & A due to overheating Slows down its speed to that of C & D but continues and doesn't stop at Clock tower, WHEN C & D Reach clock tower (1 hour after A & B had reached there) B starts its engine & C increases its speed to keep pace with B & D Continues at its usual slower speed! Finally when A reaches the END, B & C are 10 KMs behind it (Remember A had slowed down to original speeds of C & D BUT had continued traveling for that 1 hour when C & D were reaching Clock Tower & B was resting at CLock Tower BUT when C reached hen Both B & C proceeded at higher speed i.e. the speed with which A & B were traveling initially) & when B & C Reached THE END, D Was 30 Mins behind them (i.e. D took another half to reach the end!
1. What is the total distance of the Race? 2. How far was Clock tower from the Point of Start?
@bodhi_vrikshaI got 9 ...approach is a bit lengthy, so will only post the logic i used, not the entire solution.1. Took the 7 digit number mod 1001, since 1001 is a multiple of 7, and if we divide the number say abcdefg into groups of 3 and subtract the ones at even position from the ones at odd position, the final number would be 1,2 or 3 digit number which is 2. since we know g=5, we can rewrite it as (ef5+a- bcd)mod 7=0, where ef5 and cde are three digit numbers, and a is a singe digit number.3. Writing it in form of an equation (100(e-b) +10(f-c) +5+(a-d)) mod 7=04. since 100 mod 7=2 and 10 mod 7=3. we can rewrite above equation in terms of remainders alone and get (2(e-b) +13(f-c) +(5+a-d)) mod 7=05. Now since the number consists of only 5's and 7's, the values inside the brackets can only yield 3 values 0,2 and -2 respectively.6. now take cases and arrangements. eg.when 1st and 2nd bracket are 0, 5+(e-a) mod7 should give 0 and and hence the bracket value of +2 fits here. So e=b f=c, a=7, d=5. and the ea pair, and fd pair can each take 2 values of both being 5 or 7 respectively, so it can be 75555557. Continue on with the various other cases and you arrive at 9 solutions, i may have missed out a case so please confirm the OA.
I would like to say that your estimate is in the ball park range Let me create a few nice options for you a) 8 b)9 c>10 d>12 e>15 But the question is about finding those numbers and not just the number of such numbers!
9 cases, I had missed 1.
If the number is abcdefg then a +5b +4c +6d +2e +3f +g should be divisible by 7. Also g = 5. hence a +5b +4c +6d +2e +3f = 7k+2. Also the variables are either 5 or 7. Solving, we get 9 cases:
PQR is a right angled triangle. S, T and V represent the points of contact, of the incircle of the triangle PQR on side PQ, QR and RP respectively, with the sides of the triangle. PS / SQ and RT /TQ are integers (PS /SQ) + (RT / TQ) = a) 3 b) 4 c) 5 d) 6
PQR is a right angled triangle. S, T and V represent the points of contact, of the incircle of the triangle PQR on side PQ, QR and RP respectively, with the sides of the triangle. PS / SQ and RT /TQ are integers (PS /SQ) + (RT / TQ) =a) 3b) 4c) 5d) 6
PQR is a right angled triangle. S, T and V represent the points of contact, of the incircle of the triangle PQR on side PQ, QR and RP respectively, with the sides of the triangle. PS / SQ and RT /TQ are integers (PS /SQ) + (RT / TQ) =a) 3b) 4c) 5d) 6
PQR is a right angled triangle. S, T and V represent the points of contact, of the incircle of the triangle PQR on side PQ, QR and RP respectively, with the sides of the triangle. PS / SQ and RT /TQ are integers (PS /SQ) + (RT / TQ) =a) 3b) 4c) 5d) 6
Though you haven't mentioned it, but there cannot be any other case, that triangle is right angled at Q and PR is the hypotenuse.
Due to theorem of equal tangents to a circle from same exterior points, we know that
QS = QT = x (say)
And PS = PV = ax (say) where a is an integer
Also RT = RV = bx (say) where b is an integer
Applying Pythagoras in PQR, we get
(ax + x)^2 + (bx + x)^2 = (ax + bx)^2 i.e. (2a + 1) + (2b + 1) = 2ab i.e. (a - 1)(b - 1) = 2
Find the largest positive integer which is divisible by all positive integers which are less than or equal to its square root. e.g. 12 is divisible by all positive integers which are less than or equal to its square root.
One from my side now: easy one Find the largest positive integer which is divisible by all positive integers which are less than or equal to its square root.e.g. 12 is divisible by all positive integers which are less than or equal to its square root.Team BV
24?
Beyond 25 will have to be divisible by 2, 3, 4, 5 = min 60 => too large, crosses 49 as well... regards scrabbler
One from my side now: easy one Find the largest positive integer which is divisible by all positive integers which are less than or equal to its square root.e.g. 12 is divisible by all positive integers which are less than or equal to its square root.Team BV
Ooops...it became much easier :D
I intended to ask: Find the largest positive integer which is divisible by all positive integers which are less than or equal to its cube root. e.g. 120 is divisible by all positive integers which are less than or equal to its cube root.
Ooops...it became much easier I intended to ask: Find the largest positive integer which is divisible by all positive integers which are less than or equal to its cube root.e.g. 120 is divisible by all positive integers which are less than or equal to its cube root.Team BV
420 then. Divisible by all up to 7. 840 is the next, and the first which satisfies for up to 8, but doesn't work for 9 (cube root of 729).
I realised my mistake after I posted! 
