ABC26GN0232 · The number of zeros at the end of 60 ! is
ABC26GN0233 · The numbers 1, 3, 5, 7, 99 and 128 are multiplied together. The number of zeros at the end of the product must be
ABC26GN0234 · The numbers 2, 4, 6, 8, \_\_\_\_ 98, 100 are multiplied together. The number of zeros at the end of the product must be
ABC26GN0235 · Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all the elements of S. With how many consecutive…
ABC26GN0236 · Find the number of zeros at the end of the result 3 6 9 12 15 99 102.
ABC26GN0237 · The unit's digit of 132003 is · 2010
ABC26GN0238 · The digit in the unit's place of the number 123 99 is · 2007
ABC26GN0239 · Match List I with List II and select the correct answer: List I List II (Product) (Digit in the unit's place) A. (1827) 16 (1) 1 B…
ABC26GN0240 · The digit in the unit's place of the number (67)25-1 must be
ABC26GN0241 · The unit's digit in the product 274 318 577 313 is
ABC26GN0242 · In the product 459 46 28* 484, the digit in the unit place is 8. The digit to come in place of * is
ABC26GN0243 · The digit in the unit place of the number represented by (795-358 ) is
ABC26GN0244 · Unit's digit in (784)126+(784)127 is
ABC26GN0245 · The digit in the unit's place of [(251)98+(21)29- . .(106)100+(705)35-164+259 ] is
ABC26GN0246 · The digit in the unit's place of the product (2464)1793 (615)317 (131)491 is
ABC26GN0247 · If x is an even number, then x4 n, where n is a positive integer, will always have
ABC26GN0248 · If m and n are positive integers, then the digit in the unit's place of 5n+6m is always
ABC26GN0249 · The number formed from the last two digits (ones and tens) of the expression 212 n-64 n, where n is any positive integer is · 2005
ABC26GN0250 · The last digit in the decimal representation of (1/5 )2000 is (Hotel Management, 2009)
ABC26GN0251 · Let x be the product of two numbers 3,659,893,456,789,325,678 and 342,973,489,379,256. The number of digits in x is · 2010
ABC26GN0252 · Let a number of three digits have for its middle digit the sum of the other two digits. Then it is a multiple of · 2008
ABC26GN0253 · What least value must be given to n so that the number 6135 n 2 becomes divisible by 9 ? · 2008
ABC26GN0254 · Find the multiple of 11 in the following numbers. · 2006
ABC26GN0256 · Which of the following numbers is not divisible by 18?
ABC26GN0257 · The number 89715938* is divisible by 4. The unknown non-zero digit marked as * will be
ABC26GN0258 · Which one of the following numbers is divisible by 3?
ABC26GN0259 · A number is divisible by 11 if the difference between the sums of the digits in odd and even places respectively is
ABC26GN0260 · Which one of the following numbers is divisible by 11?
ABC26GN0261 · Which one of the following numbers is divisible by 15?
ABC26GN0263 · Consider the following statements: The numbers 24984, 26784 and 28584 are (1) divisible by 3 (2) divisible by 4 (3) divisible by 9 Which of…
ABC26GN0264 · Which of the following numbers is a multiple of 8 ?
ABC26GN0265 · If 78*3945 is divisible by 11, where * is a digit, then * is equal to
ABC26GN0266 · If m and n are integers divisible by 5, which of the following is not necessarily true?
ABC26GN0267 · An integer is divisible by 16 if and only if its last X digits are divisible by 16. The value of X would be
ABC26GN0268 · Which of the following numbers is divisible by 3, 7, 9 and 11?
ABC26GN0269 · A number 476**0 is divisible by both 3 and 11. The non-zero digits in the hundred's and ten's place respectively are
ABC26GN0270 · How many of the following numbers are divisible by 3 but not by 9? 2133, 2343, 3474, 4131, 5286, 5340, 6336, 7347, 8115, 9276
ABC26GN0271 · If the number 357*25* is divisible by both 3 and 5, then the missing digits in the unit's place and the thousandth's place respectively are
ABC26GN0275 · 325325 is a six-digit number. It is divisible by
ABC26GN0276 · If the seven-figure number 30X0103 is a multiple of 13, then X is
ABC26GN0277 · If a number is divisible by both 11 and 13, then it must be necessarily
ABC26GN0278 · Which of the following numbers are completely divisible by 7? I. 195195 II. 181181 III. 120120 IV. 891891 self-practice
ABC26GN0279 · If x and y are two digits of the number 653 x y such that the number is divisible by 80, then x+y is equal to
ABC26GN0280 · The six-digit number 5 A B B 7 A is a multiple of 33 for non-zero digits A and B. Which of the following could be possible value of A+B ? · 2010
ABC26GN0281 · Which of the following numbers is divisible by 99?
ABC26GN0282 · The digits indicated by * in 3422213** so that this number is divisible by 99 are · 2010