Number System: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 5)

Let n = 93. The sum of the first n natural numbers is given by the formula n(n+1)/2. Substituting n = 93: sum = 93 x (93+1)/2 = 93 x 94/2 = 8742/2 = 4371. Check: for a small case like n=3, the formula gives 3×4/2=6, which matches 1+2+3=6, confirming the formula is applied correctly.

Let n = 100. The sum of the first n natural numbers is given by the formula n(n+1)/2. Substituting n = 100: sum = 100 x (100+1)/2 = 100 x 101/2 = 10100/2 = 5050. Check: for a small case like n=3, the formula gives 3×4/2=6, which matches 1+2+3=6, confirming the formula is applied correctly.

Let n = 167. The sum of the first n natural numbers is given by the formula n(n+1)/2. Substituting n = 167: sum = 167 x (167+1)/2 = 167 x 168/2 = 28056/2 = 14028. Check: for a small case like n=3, the formula gives 3×4/2=6, which matches 1+2+3=6, confirming the formula is applied correctly.

Let n = 199. The sum of the first n natural numbers is given by the formula n(n+1)/2. Substituting n = 199: sum = 199 x (199+1)/2 = 199 x 200/2 = 39800/2 = 19900. Check: for a small case like n=3, the formula gives 3×4/2=6, which matches 1+2+3=6, confirming the formula is applied correctly.

Let n = 118. The sum of the first n natural numbers is given by the formula n(n+1)/2. Substituting n = 118: sum = 118 x (118+1)/2 = 118 x 119/2 = 14042/2 = 7021. Check: for a small case like n=3, the formula gives 3×4/2=6, which matches 1+2+3=6, confirming the formula is applied correctly.

Let n = 185. The sum of the first n natural numbers is given by the formula n(n+1)/2. Substituting n = 185: sum = 185 x (185+1)/2 = 185 x 186/2 = 34410/2 = 17205. Check: for a small case like n=3, the formula gives 3×4/2=6, which matches 1+2+3=6, confirming the formula is applied correctly.

Let n = 165. The sum of the first n natural numbers is given by the formula n(n+1)/2. Substituting n = 165: sum = 165 x (165+1)/2 = 165 x 166/2 = 27390/2 = 13695. Check: for a small case like n=3, the formula gives 3×4/2=6, which matches 1+2+3=6, confirming the formula is applied correctly.

Let n = 62. The sum of the first n natural numbers is given by the formula n(n+1)/2. Substituting n = 62: sum = 62 x (62+1)/2 = 62 x 63/2 = 3906/2 = 1953. Check: for a small case like n=3, the formula gives 3×4/2=6, which matches 1+2+3=6, confirming the formula is applied correctly.

We need multiples of 4 between 300 and 1199 inclusive. The first multiple of 4 at or above 300 is 300. The last multiple of 4 at or below 1199 is 1196. Count = (last – first)/4 + 1 = (1196-300)/4 + 1 = 225. So there are 225 such numbers.

We need multiples of 13 between 200 and 1099 inclusive. The first multiple of 13 at or above 200 is 208. The last multiple of 13 at or below 1099 is 1092. Count = (last – first)/13 + 1 = (1092-208)/13 + 1 = 69. So there are 69 such numbers.

We need multiples of 12 between 300 and 1199 inclusive. The first multiple of 12 at or above 300 is 300. The last multiple of 12 at or below 1199 is 1188. Count = (last – first)/12 + 1 = (1188-300)/12 + 1 = 75. So there are 75 such numbers.

We need multiples of 13 between 100 and 999 inclusive. The first multiple of 13 at or above 100 is 104. The last multiple of 13 at or below 999 is 988. Count = (last – first)/13 + 1 = (988-104)/13 + 1 = 69. So there are 69 such numbers.

We need multiples of 8 between 200 and 1099 inclusive. The first multiple of 8 at or above 200 is 200. The last multiple of 8 at or below 1099 is 1096. Count = (last – first)/8 + 1 = (1096-200)/8 + 1 = 113. So there are 113 such numbers.

We need multiples of 12 between 100 and 999 inclusive. The first multiple of 12 at or above 100 is 108. The last multiple of 12 at or below 999 is 996. Count = (last – first)/12 + 1 = (996-108)/12 + 1 = 75. So there are 75 such numbers.

We need multiples of 7 between 300 and 1199 inclusive. The first multiple of 7 at or above 300 is 301. The last multiple of 7 at or below 1199 is 1197. Count = (last – first)/7 + 1 = (1197-301)/7 + 1 = 129. So there are 129 such numbers.

We need multiples of 8 between 300 and 1199 inclusive. The first multiple of 8 at or above 300 is 304. The last multiple of 8 at or below 1199 is 1192. Count = (last – first)/8 + 1 = (1192-304)/8 + 1 = 112. So there are 112 such numbers.

We need multiples of 9 between 300 and 1199 inclusive. The first multiple of 9 at or above 300 is 306. The last multiple of 9 at or below 1199 is 1197. Count = (last – first)/9 + 1 = (1197-306)/9 + 1 = 100. So there are 100 such numbers.

We need multiples of 3 between 200 and 1099 inclusive. The first multiple of 3 at or above 200 is 201. The last multiple of 3 at or below 1099 is 1098. Count = (last – first)/3 + 1 = (1098-201)/3 + 1 = 300. So there are 300 such numbers.

The available digits are 0, 9, 5, 2, 4. To minimise a number, the leading digit must be non-zero (a leading zero would drop the digit count), so the smallest non-zero digit, 2, leads. The remaining digits [0, 4, 5, 9] then follow in ascending order. This gives 20459. So the minimum 5-digit number is 20459.

The available digits are 8, 4, 0, 7, 2. To maximise a number, its digits should go in descending order, from largest to smallest. Arranging 8, 4, 0, 7, 2 in descending order gives [8, 7, 4, 2, 0], which reads as 87420. So the maximum 5-digit number is 87420.

The available digits are 0, 7, 9, 5, 8. To maximise a number, its digits should go in descending order, from largest to smallest. Arranging 0, 7, 9, 5, 8 in descending order gives [9, 8, 7, 5, 0], which reads as 98750. So the maximum 5-digit number is 98750.

The available digits are 4, 3, 8, 5, 0. To minimise a number, the leading digit must be non-zero (a leading zero would drop the digit count), so the smallest non-zero digit, 3, leads. The remaining digits [0, 4, 5, 8] then follow in ascending order. This gives 30458. So the minimum 5-digit number is 30458.

The available digits are 0, 2, 3, 5. To maximise a number, its digits should go in descending order, from largest to smallest. Arranging 0, 2, 3, 5 in descending order gives [5, 3, 2, 0], which reads as 5320. So the maximum 4-digit number is 5320.

The available digits are 6, 4, 0, 1. To minimise a number, the leading digit must be non-zero (a leading zero would drop the digit count), so the smallest non-zero digit, 1, leads. The remaining digits [0, 4, 6] then follow in ascending order. This gives 1046. So the minimum 4-digit number is 1046.

The available digits are 9, 6, 0, 4. To minimise a number, the leading digit must be non-zero (a leading zero would drop the digit count), so the smallest non-zero digit, 4, leads. The remaining digits [0, 6, 9] then follow in ascending order. This gives 4069. So the minimum 4-digit number is 4069.