R(n!) = Last Digit of [ 2a x R(a!) x R(b!) ] where n = 5a + b
Eg1.1: What is the rightmost non-zero digit of 37! ? R (37!) = Last Digit of [ 2^7 x R (7!) x R (2!) ] R (37!) = Last Digit of [ 8 x 4 x 2 ] = 4
Eg 1.2: What is the rightmost non-zero digit of 134! ? R (134!) = Last Digit of [ 2^26 x R (26!) x R (4!) ] R (134!) = Last Digit of [ 4 x R (26!) x 4 ] We need to find out R (26!) = Last Digit of [ 2^5 x R (5!) x R (1!) ] = Last digit of [ 2 x 2 x 1 ] = 4 So, R (134!) = Last Digit of [ 4 x 4 x 4 ] = 4
@chandrakant.k@anwesa09 i found one approach-please review the same and provide the feedback
Last two non zero digit of 38!? N=38!. 38 = 1*2*3*4*5*6*7*8*9*10 11*12*13*14*15*16*17*18*19*20 21*22*23*24*25*26*27*28*29*30 31*32*33*34*35*36*37*38 Lets remove all the zeroes. Number of 0s to be removed = Highest power of 5 that divides 38! = 7 + 1 = 8. We will first remove the 5s. So now 38/5^8 becomes = 1*2*3*4*1*6*7*8*9*2 11*12*13*14*3*16*17*18*19*4 21*22*23*24*1*26*27*28*29*6 31*32*33*34*7*36*37*38 Now we that product of 1*2*3*4*6*7*8*9 ends with 76. So does 11*12..*19 and 21*22..*29.. For 31*32*..38, you can either multiply the last 2 digits or do the following. we know that 31*32*..*39 ends with 76. So let 31*32..*38 end with ab. 76 = (ab * 39)%100. b = 4., 9a + 15 = 7 => a = 8. So 31*32..*38 = 84. After removing all the 5s, we collect 1*2*3..*8*9 separately(lets call it A) and collect the rest separately. The rest is (1,2,3,4,6,7). Lets call it B. A = 76 *76* 76 * 84 = 76*84 = 84 B = 08 So A * B mod 100 = 84 * 8 mod 100 = 72 Now, what we really need is last 2 digits of 38! / 10^8. So we need to remove 2^8 from the above. Instead, we can do the following. Let ab be the last 2 digits of 38! /10^8. We know that 38!/5^8 ends with 72. So if we multiple ab by 2^8, we should get 72. 2^8 mod 100 = 56. 72 = ab * 56 b = 2 or 7. But N! cant end in odd non-zero digit. Hence b=2. (6a +11) mod 10 = 7, a= 1, 6 Hence it can end in 12 or 62. Now 38! is divisible by 4. So it must end in 12 and not 62. Though the method looks long, you need not enumerate all the above numbers in column format. It can be shortened to 4-5 lines
16 days...convert the work into manhr..i.e. 10men for 9 days = 90 manhr6women for 9 days = 6*7.5 = 45womenhr = 45*2/3(to convert their equivalent effort into a man's effort) = 30 manhrtotal manhr for 1 day = 120manhrfor 18 days = 120*182nd case :10men *9+ 9*7.5*2/3 = 135manhr120*18 = 135*XX= 16 days...whats the OA??
Q. A and B start moving towards each other simultaneously on a straight line from cities P and Q respectively. After travelling some distance, B takes a 30Δ§¸ turn to his left with respect to his original direction. 2 hour after B turns, A takes a 90Δ§¸ turn to his right. A travels 60 km after turning, before meeting B. They meet 10 hour after starting their journey. A and B together travel 170 km with time ratio 8:9 respectively before turning and arrive at the meeting point simultaneously. If A and B had not turned, after how many hours would they have met?a) 34 +(12root3)/7b)34+ (8root3)/7c)34+ (16root3)/7D)34 + (12root3)/5
@joyjitpal please explain how come it will be more than 10 hours in option. as in the question itself it is mentioned that they meet 10 hour after starting their journey. if they are not turning it has to be less than 10 hours.
A is twice as efficient as B and together they do the same work in as much time as C and D. If C and D can complete the work in 20 and 30 days respectively working alone, thn in how many days A can complete the work individually?
A is twice as efficient as B and together they do the same work in as much time as C and D. If C and D can complete the work in 20 and 30 days respectively working alone, thn in how many days A can complete the work individually?
A is twice as efficient as B and together they do the same work in as much time as C and D. If C and D can complete the work in 20 and 30 days respectively working alone, thn in how many days A can complete the work individually?
A is twice as efficient as B and together they do the same work in as much time as C and D. If C and D can complete the work in 20 and 30 days respectively working alone, thn in how many days A can complete the work individually?
Among 4 people A,B,C and D, A takes thrice as much as B to complete a piece of work. B takes thrice as much time as C and C takes thrice as much time as D to complete the same work. One group of three of the 4 men can complete the work in 13 days while another group of three can do so in 31 days. Which is the group that takes 13 days? a) A,B,D b)B,C,D c)A,C,D d)A,B,C
177/3= 59 =abc. say a=59, bc=1b+c>=2 sqrt(bc)or b+c>=2 (so min value of b+c=2)min val of exp = (59+2)*(59(b+c) +1)= 61*(119).(are all the values quoted in your post correct? OA doesnt seem to match!)
yes. all the values are correct. u cant suppose the values of a, bc like that.
A is twice as efficient as B and together they do the same work in as much time as C and D. If C and D can complete the work in 20 and 30 days respectively working alone, thn in how many days A can complete the work individually?
B=x days A takes x/2 days
together they do it in x/3 days
which is equivalent to C+ D working together. C+D do the work in 12 days. so x/3=12 x=36
@ananyboss said:What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... (1152!)^3 is divided by 1152. ??Ans- Approach Please, 1152 = 2^7*3^2 leave 1!^3 = 1 2!^3= 8 3!^3= (2^3)*(3^3) 4! ^3= (2^9)*(3^3)... so 1152 will divide this and rem from this onward should be 0. calculate for 1st 3 terms only. (1+8+216)/1152= 225 ...is rem. plz correct me if i m wrong.