ok... my final solution to this problem we need last non-zero digit of 1786! this s last digit of 1780!*6!(consider only last digits of 1781,....) last non zero digit of 6!=2 now last digit of 1.2.3.4.5.6.7.8.9=8 this occurs 178 times so we need to find last digit of 8^178 but here we havent accounted for the digit changes by 20,30,40,50,60,.....till 1780 so we need to multiply by 178! hence we need last non zero digit of 178!*2*8^178 last digit of 8^178=last digit of 8^2=4 last digit of 178!=last digit of 8^170*8!*17! =last digit of 4*8!*17! =last digit of 4*8!*10!*7! =2*4*2=6
hence last nonzero digit is 6*2*4 =8 aaarghhh!! wth is the mistake!!!?? :banghead:
Juz a clue to u...last non zero digit of 14 is 4 , then for 14*15 is???? 4 or 0:biggrin? No it's 1 (15*14 is 210)and this 5 changes the whole scenario. so u can't apply this concept here!!
Q> How many integers less than 300 are relatively prime to either 10 or 18?
1. 140 2. 141 3. 142 4. 139 5. 138
Please post the approach also
i tried using set theory funda n (a intersection b) = n(a)+ n(b) - n (a u b)
n(a) = no.s divisible by 10, 300/10 = 30 n(b) = " '" '" 18 , 300/18= 16 n (a u b) = " " by both a and b = nos divisible by lcm of 10 & 18 = 90 , n (a u b) = 3
so n (a intersection b) = 30+16-3 = 43 ... but this is not matching with the answers you provided some veteran please check my approach
i tried using set theory funda n (a intersection b) = n(a)+ n(b) - n (a u b)
n(a) = no.s divisible by 10, 300/10 = 30 n(b) = " '" '" 18 , 300/18= 16 n (a u b) = " " by both a and b = nos divisible by lcm of 10 & 18 = 90 , n (a u b) = 3
so n (a intersection b) = 30+16-3 = 43 ... but this is not matching with the answers you provided some veteran please check my approach
numbers less than 300 and co-prime to 2 or 5 is : 300 - - + = 120 numbers less than 300 and co-prime to 2 or 3 is : 300 - - + = 100 numbers less than 300 and co-prime to 2, 3 and 5 is : 300 - - - + - - - = 80 Therefore for all the possible integers less than 300 relatively prime to either 10 or 18 are: So, 120+100-80 = 140.
Directions for next 3 questions: Each question is followed by two statements, A and B. Answer each question using the following instructions:
Choose (1) if the question can be answered by one of the statements alone, but cannot be answered by using the other statement alone. Choose (2) if the question can be answered by using either statement alone. Choose (3) if the question can be answered by using both the statements together, but cannot be answered by using either statement alone. Choose (4) if the question cannot be answered even by using both statements together.
1. A alone can complete the task in 10 days and B alone in 12 days. On which day does the task get completed? A. A and B take turns, working on alternate days. B. On the 3rd day it is A's turn a. 1 b. 2 c. 3 d. 4
2. Which is the smallest of the three positive integers a, b and c? A. ab = 6 B. bc = 12, ac = 8 a. 1 b. 2 c. 3 d. 4
3. The income of (A) and (B) are in the ratio 5:3 and their expenses in the ratio 9:4. Find their incomes. A. Each of them saves Rs. 700. B. Each of them saves equal amounts. a. 1 b. 2 c. 3 d. 4
hey jain_ashu could you explain with another example? just to make it more clear
Sorry for late reply. What I have done is factorize the numbers 10 and 18 into their prime factors. for 10 it is 2 and 5 and for 18 it is 2 and 3. After that its basically a set theory concept.
Directions for next 3 questions: Each question is followed by two statements, A and B. Answer each question using the following instructions:
Choose (1) if the question can be answered by one of the statements alone, but cannot be answered by using the other statement alone. Choose (2) if the question can be answered by using either statement alone. Choose (3) if the question can be answered by using both the statements together, but cannot be answered by using either statement alone. Choose (4) if the question cannot be answered even by using both statements together.
1. A alone can complete the task in 10 days and B alone in 12 days. On which day does the task get completed? A. A and B take turns, working on alternate days. B. On the 3rd day it is A's turn a. 1 b. 2 c. 3 d. 4
2. Which is the smallest of the three positive integers a, b and c? A. ab = 6 B. bc = 12, ac = 8 a. 1 b. 2 c. 3 d. 4
3. The income of (A) and (B) are in the ratio 5:3 and their expenses in the ratio 9:4. Find their incomes. A. Each of them saves Rs. 700. B. Each of them saves equal amounts. a. 1 b. 2 c. 3 d. 4
1. The answer is option (1) as no matter in which order they do work, it gets completed on the 6th day. 2. The answer would be Option (3) as both the sets of equations are required to get all the 3 values. 3. The answer would be option (2) , but only if the question asked to find the ratio of their savings and not their incomes...
Q> How many integers less than 300 are relatively prime to either 10 or 18?
1. 140 2. 141 3. 142 4. 139 5. 138
Please post the approach also
i tried using set theory funda n (a intersection b) = n(a)+ n(b) - n (a u b)
n(a) = no.s divisible by 10, 300/10 = 30 n(b) = " '" '" 18 , 300/18= 16 n (a u b) = " " by both a and b = nos divisible by lcm of 10 & 18 = 90 , n (a u b) = 3
so n (a intersection b) = 30+16-3 = 43 ... but this is not matching with the answers you provided some veteran please check my approach
This can be done using Euler number concept.........
As below 300 no. coprime to 10 are -> 300*(1/2)*(4/5) = 120
Below 300 coprime to 18 are -> 300*(1/2)*(2/3) = 100
Below 300 coprime to 300 -> 300*(1/2)*(2/3)*(4/5) = 80
Directions for next 3 questions: Each question is followed by two statements, A and B. Answer each question using the following instructions:
Choose (1) if the question can be answered by one of the statements alone, but cannot be answered by using the other statement alone. Choose (2) if the question can be answered by using either statement alone. Choose (3) if the question can be answered by using both the statements together, but cannot be answered by using either statement alone. Choose (4) if the question cannot be answered even by using both statements together.
1. A alone can complete the task in 10 days and B alone in 12 days. On which day does the task get completed? A. A and B take turns, working on alternate days. B. On the 3rd day it is A's turn a. 1 b. 2 c. 3 d. 4
ans for this shud be: 1 as in this case it will not matter who is starting it first both the cases will end up finishing the work in 10 days..
Directions for next 3 questions: Each question is followed by two statements, A and B. Answer each question using the following instructions:
Choose (1) if the question can be answered by one of the statements alone, but cannot be answered by using the other statement alone. Choose (2) if the question can be answered by using either statement alone. Choose (3) if the question can be answered by using both the statements together, but cannot be answered by using either statement alone. Choose (4) if the question cannot be answered even by using both statements together.
2. Which is the smallest of the three positive integers a, b and c? A. ab = 6 B. bc = 12, ac = 8 a. 1 b. 2 c. 3 d. 4
answer for this shud be 3.. as using both the statements we will get the answer as a=2 b=3 c=4..
if a^2 +b^2=1553 and a,b are integers,the remainder when (a+b) is divided by 7 cannot be: 1) 1 2)2 3)4 4)5, I understand both the integers must be less than 40,please explain the method:-o
if a^2 +b^2=1553 and a,b are integers,the remainder when (a+b) is divided by 7 cannot be: 1) 1 2)2 3)4 4)5, I understand both the integers must be less than 40,please explain the method:-o
For such questions, familiarity with squares is of great importance. You need to get at the values of 'a' and 'b' by hit and trial. In this case, the values are 1024 and 529. Since squares are in the picture, you need to remember to take positive as well as negative values. here, a will be + 23 and b will be + 32. Now work out the possible combinations and check for the required divisibility.
if a^2 +b^2=1553 and a,b are integers,the remainder when (a+b) is divided by 7 cannot be: 1) 1 2)2 3)4 4)5, I understand both the integers must be less than 40,please explain the method:-o
As Gupta Ji Has Already Posted Above...My Approach Unit's Digit 3 can be formed only and only when Sum of Digits of squares is ending with 4 and 9....
So, Only Squares which can be possible together are + 32 and + 23. Combinations = 32 - 23 = 9 mod 7 = 2 32+23 mod 7 = 55 mod 7 = 6 -32-23 mod 7 = -55 mod 7 = 1 -32+23 mod 7 = -9 mod 7 = 5 Answer is 4
i have a question......in how many ways 1 billion( 1,000,000,000) be written as a product of two numbers such that neither of the nos is a multiple of 10 ?
i have a question......in how many ways 1 billion( 1,000,000,000) be written as a product of two numbers such that neither of the nos is a multiple of 10 ?
This can be done using Euler number concept.........
As below 300 no. coprime to 10 are -> 300*(1/2)*(4/5) = 120
Below 300 coprime to 18 are -> 300*(1/2)*(2/3) = 100
Below 300 coprime to 300 -> 300*(1/2)*(2/3)*(4/5) = 80
So Required ans = 120 + 100 - 80 = 140
what are the various uses of euler's rule, i know the one about remainders and i can see the same funda is being applied here too but couldn't understand the concept exactly, could you tell me where you found out this concept from some website etc ; otherwise please explain in your own words
can smebdy tell me the concepts of finding ten's digit??? Like ... Find 2^1001 ten's digit ?? Please let me knw all possible methods.......n explain dem...:cheers:
can smebdy tell me the concepts of finding ten's digit??? Like ... Find 2^1001 ten's digit ?? Please let me knw all possible methods.......n explain dem...:cheers:
? No it's 1 (15*14 is 210)and this 5 changes the whole scenario. so u can't apply this concept here!!
some veteran please check my approach 
:cheers: