The sum of 1/4,1/9,1/16....1/n2 is :
(1)	less than 1	(2)	greater than 1
(3)	equal to 1 or less than 1	(4)	equal to 1/n2 (n + 2)
The right answer is Option (1).
¼ + 1/9 + 1/16 +…+ 1/n2 Consider ¼: ¼ = 1/(2 x 2) < 1/(1 x 2) because ½ < 1 Similarly, 1/9 = 1/(3 x 3) < 1/(2 x 3) Since 1/3 < ½ 1/16 = 1/(4 x 4) < 1/(3 x 4) Because ¼ < 1/3 …………………………………………………….. …………………………………………………….. 1/n2 = 1/(nxn) < 1/(n – 1)n Since 1/n < 1/(n – 1) ¼ + 1/9 + …+ 1/n2 < 1/(1 x 2) + 1/(2 x 3) + … + 1/[(n – 1)n] i.e. ¼ + 1/9 + 1/16 + … + 1/n2 < (1/1 – ½) + (1/2 – 1/3) +…+ [1/(n – 1) – 1/n] i.e. ¼ + 1/9 + 1/16 +… + 1/n2 < 1/1 – 1/n which is less than 1. Hence, the given sum is less than 1 only. Answer: (1) What is the digit in the unit place of  3946252 + 734733 ? (1) 8 (2)  7 (3) 6 (4) 4 The right answer is Option (1). 3946252 + 734733 First we will find the unit digit of 3946252. Divide 46, by 4, we get 2 as remainder. Actual remainder (2)252 Again dividing (2)252  by 4, we get 0 as remainder. And ‘9’ is an odd digit number. So, the last digit of 3946252 will be 1. Last digit of 734733 Þ By dividing 47 by 4, we get 3 or – 1 as a remainder. Now, (– 1) odd, when divided by 4 will give – 1 or 3 as remainder. So, the last digit of 734733 will be. Last digit of 3 ´ 3 ´ 3 = i.e. 7. So, the last digit is 1 + 7 = 8. Answer: (1) The expression 333555 + 555333 is divisible by: (1) 2 (2)  3 (3) 37 (4) All of these The right answer is Option (4). 333555 + 555333 Check it by option. The last digit of 333555 will be odd, and the last digit of 555333 will be odd. odd + odd Þ even So, the last digit is even, which is divisible by 2. Both the numbers are individually divisible by 3. So 333555 + 555333 is divisible by 3. And similarly, both the numbers are divisible by 37, so, 333555 + 555333 is divisible by 37. Hence this expression is divisible by all three numbers.   Answer : (4) Note : If the number consists of same in the multiples of 3, then it is always a multiple of 37 � l ; p� �ߡ ight:150%;font-family:"Arial","sans-serif";color:#333333'> So, the last digit is 1 + 7 = 8. Answer: (1)

divisiable rule

Number	Method	Example
7	Subtract 2 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 7, the original number is also divisible by 7	Check for 945: : 94-(2*5)=84. Since 84 is divisible by 7, the original no. 945 is also divisible
13	Add 4 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 13, the original number is also divisible by 13	Check for 3146:: 314+ (4*6) = 338:: 33+(4*8) = 65. Since 65 is divisible by 13, the original no. 3146 is also divisible
17	Subtract 5 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 17, the original number is also divisible by 17	Check for 2278:: 227-(5*8)=187. Since 187 is divisible by 17, the original number 2278 is also divisible.
19	Add 2 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 19, the original number is also divisible by 19	Check for 11343:: 1134+(2*3)= 1140. (Ignore the 0):: 11+(2*4) = 19. Since 19 is divisible by 19, original no. 11343 is also divisible
23	Add 7 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 23, the original number is also divisible by 23	Check for 53935:: 5393+(7*5) = 5428 :: 542+(7*8)= 598:: 59+ (7*8)=115, which is 5 times 23. Hence 53935 is divisible by 23
29	Add 3 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 29, the original number is also divisible by 29	Check for 12528:: 1252+(3*8)= 1276 :: 127+(3*6)= 145:: 14+ (3*5)=29, which is divisible by 29. So 12528 is divisible by 29
31	Subtract 3 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 31, the original number is also divisible by 31	Check for 49507:: 4950-(3*7)=4929 :: 492-(3*9) :: 465:: 46-(3*5)=31. Hence 49507 is divisible by 31
37	Subtract 11 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 37, the original number is also divisible by 37	Check for 11026:: 1102 - (11*6) =1036. Since 103 - (11*6) =37 is divisible by 37. Hence 11026 is divisible by 37
41	Subtract 4 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 41, the original number is also divisible by 41	Check for 14145:: 1414 - (4*5) =1394. Since 139 - (4*4) =123 is divisible by 41. Hence 14145 is divisible by 41
43	Add 13 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 43, the original number is also divisible by 43.*This process becomes difficult for most of the people because of multiplication with 13.	Check for 11739:: 1173+(13*9)= 1290:: 129 is divisible by 43. 0 is ignored. So 11739 is divisible by 43
47	Subtract 14 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 47, the original number is also divisible by 47. This too is difficult to operate for people who are not comfortable with table of 14.	Check for 45026:: 4502 - (14*6) =4418. Since 441 - (14*8) =329, which is 7 times 47. Hence 45026 is divisible by 47

Notes:

Isro's 100th mission: PSLV-C21 puts 2 foreign satellites in orbit
CHENNAI: The Indian space odyssey crossed a historic landmark on Sunday when a PolarSatellite Launch Vehicle (PSLV-C21) put in orbit two foreign satellites.

It marked the 100th space mission of the Indian Space Research Organisation (Isro) which started the journey in 1975 with the launch of its first satellite 'Aryabhata.'

Among those who watched the majestic rocket lift off from the Satish Dhawan Space Centre in Sriharikota, 100km north of Chennai, was Prime Minister Manmohan Singh.

At the end of a 51-hour countdown that started on Friday, PSLV-C21 lifted off at 9.53am. After 17 minutes and 49 seconds, it injected the first satellite, the French SPOT-6, into orbit. Soon, the second satellite, the Japenese Proiteres was put in orbit. There were, however, some anxious moments when the trajectory on the giant screens at the mission control room showed the rocket's blip deviate slightly midflight, which appeared to have been corrected later.

Scientists cheered and shook hands as the mission was pronounced a success and the Prime Minister congratulated the scientists. "India is justly proud of its space scientists who have overcome many odds to develop our own technology for space missions," he said.

It was PSLV's 21st consecutively successful flight. PSLV-C21 is India's 38th satellite launch vehicle to lift off from Sriharikota. India has so far put in orbit 62 Indian satellites. The two numbers add up the century that Isro celebrated on Sunday.

Besides buttressing its technological mastery, Isro, with its latest launch, has cemented its place in among the space-faring nations as a sought-after commercial launcher. Isro's commercial arm Antrix Corporation has received several requests from foreign countries to launch their satellites for a price.