Number System: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 6)

The available digits are 0, 7, 1, 3. 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, 3, 7] then follow in ascending order. This gives 1037. So the minimum 4-digit number is 1037.

The available digits are 8, 7, 0, 2. To maximise a number, its digits should go in descending order, from largest to smallest. Arranging 8, 7, 0, 2 in descending order gives [8, 7, 2, 0], which reads as 8720. So the maximum 4-digit number is 8720.

The available digits are 2, 6, 7, 4, 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, 2, leads. The remaining digits [0, 4, 6, 7] then follow in ascending order. This gives 20467. So the minimum 5-digit number is 20467.

Let N = 20693017 and D = 9. Place value of D = D x 10000 = 9 x 10000 = 90000. Face value of D = D = 9 (a digit’s face value is always just the digit itself). Difference = place value – face value = 90000 – 9 = 89991.

Let N = 71346442 and D = 1. Place value of D = D x 1000000 = 1 x 1000000 = 1000000. Face value of D = D = 1 (a digit’s face value is always just the digit itself). Difference = place value – face value = 1000000 – 1 = 999999.

Let N = 24088184 and D = 1. Place value of D = D x 100 = 1 x 100 = 100. Face value of D = D = 1 (a digit’s face value is always just the digit itself). Difference = place value – face value = 100 – 1 = 99.

Let N = 97541648 and D = 4. Place value of D = D x 10000 = 4 x 10000 = 40000. Face value of D = D = 4 (a digit’s face value is always just the digit itself). Difference = place value – face value = 40000 – 4 = 39996.

Let N = 6163435 and D = 6. Place value of D = D x 1000000 = 6 x 1000000 = 6000000. Face value of D = D = 6 (a digit’s face value is always just the digit itself). Difference = place value – face value = 6000000 – 6 = 5999994.

Let N = 7136384 and D = 3. Place value of D = D x 100 = 3 x 100 = 300. Face value of D = D = 3 (a digit’s face value is always just the digit itself). Difference = place value – face value = 300 – 3 = 297.

Let N = 28220934 and D = 8. Place value of D = D x 1000000 = 8 x 1000000 = 8000000. Face value of D = D = 8 (a digit’s face value is always just the digit itself). Difference = place value – face value = 8000000 – 8 = 7999992.

Let N = 72037032 and D = 7. Place value of D = D x 10000000 = 7 x 10000000 = 70000000. Face value of D = D = 7 (a digit’s face value is always just the digit itself). Difference = place value – face value = 70000000 – 7 = 69999993.

Let N = 27338862 and D = 8. Place value of D = D x 100 = 8 x 100 = 800. Face value of D = D = 8 (a digit’s face value is always just the digit itself). Difference = place value – face value = 800 – 8 = 792.

Let N = 36622922 and D = 3. Place value of D = D x 10000000 = 3 x 10000000 = 30000000. Face value of D = D = 3 (a digit’s face value is always just the digit itself). Difference = place value – face value = 30000000 – 3 = 29999997.

Let N = 4418747 and D = 4. D appears 3 time(s) in N. Each occurrence has its own place value, since the same digit means different place values at different positions. Adding all 3 of D’s place values together gives 4400040. So the required sum is 4400040.

Let N = 3121499 and D = 9. D appears 2 time(s) in N. Each occurrence has its own place value, since the same digit means different place values at different positions. Adding all 2 of D’s place values together gives 99. So the required sum is 99.

Let N = 144269 and D = 4. D appears 2 time(s) in N. Each occurrence has its own place value, since the same digit means different place values at different positions. Adding all 2 of D’s place values together gives 44000. So the required sum is 44000.

Let N = 1392114 and D = 1. D appears 3 time(s) in N. Each occurrence has its own place value, since the same digit means different place values at different positions. Adding all 3 of D’s place values together gives 1000110. So the required sum is 1000110.

Let N = 861061 and D = 1. D appears 2 time(s) in N. Each occurrence has its own place value, since the same digit means different place values at different positions. Adding all 2 of D’s place values together gives 1001. So the required sum is 1001.

Let N = 8526863 and D = 8. D appears 2 time(s) in N. Each occurrence has its own place value, since the same digit means different place values at different positions. Adding all 2 of D’s place values together gives 8000800. So the required sum is 8000800.

Let N = 9354763 and D = 3. D appears 2 time(s) in N. Each occurrence has its own place value, since the same digit means different place values at different positions. Adding all 2 of D’s place values together gives 300003. So the required sum is 300003.

Let N = 165474 and D = 4. D appears 2 time(s) in N. Each occurrence has its own place value, since the same digit means different place values at different positions. Adding all 2 of D’s place values together gives 404. So the required sum is 404.

Let N = 6032630 and D = 3. D appears 2 time(s) in N. Each occurrence has its own place value, since the same digit means different place values at different positions. Adding all 2 of D’s place values together gives 30030. So the required sum is 30030.

Let N = 1276272 and D = 2. D appears 3 time(s) in N. Each occurrence has its own place value, since the same digit means different place values at different positions. Adding all 3 of D’s place values together gives 200202. So the required sum is 200202.

Let N = 23217234. The digit in question is D = 2. Counting from the right, D sits 3 place(s) away, so it occupies the 100’s place. Place value = D x (place multiplier) = 2 x 100 = 200. So the place value of 2 in 23217234 is 200.

Let N = 36068124. The digit in question is D = 1. Counting from the right, D sits 3 place(s) away, so it occupies the 100’s place. Place value = D x (place multiplier) = 1 x 100 = 100. So the place value of 1 in 36068124 is 100.

Let N = 27608544. The digit in question is D = 5. Counting from the right, D sits 3 place(s) away, so it occupies the 100’s place. Place value = D x (place multiplier) = 5 x 100 = 500. So the place value of 5 in 27608544 is 500.

Let N = 58076406. The digit in question is D = 5. Counting from the right, D sits 8 place(s) away, so it occupies the 10000000’s place. Place value = D x (place multiplier) = 5 x 10000000 = 50000000. So the place value of 5 in 58076406 is 50000000.

Let N = 25095030. The digit in question is D = 5. Counting from the right, D sits 4 place(s) away, so it occupies the 1000’s place. Place value = D x (place multiplier) = 5 x 1000 = 5000. So the place value of 5 in 25095030 is 5000.

Let N = 93083351. The digit in question is D = 9. Counting from the right, D sits 8 place(s) away, so it occupies the 10000000’s place. Place value = D x (place multiplier) = 9 x 10000000 = 90000000. So the place value of 9 in 93083351 is 90000000.

Let N = 81324990. The digit in question is D = 8. Counting from the right, D sits 8 place(s) away, so it occupies the 10000000’s place. Place value = D x (place multiplier) = 8 x 10000000 = 80000000. So the place value of 8 in 81324990 is 80000000.

Let N = 29399555. The digit in question is D = 9. Counting from the right, D sits 7 place(s) away, so it occupies the 1000000’s place. Place value = D x (place multiplier) = 9 x 1000000 = 9000000. So the place value of 9 in 29399555 is 9000000.

Let N = 52638999. The digit in question is D = 9. Counting from the right, D sits 3 place(s) away, so it occupies the 100’s place. Place value = D x (place multiplier) = 9 x 100 = 900. So the place value of 9 in 52638999 is 900.

Let N = 64866173. The digit in question is D = 6. Counting from the right, D sits 4 place(s) away, so it occupies the 1000’s place. Place value = D x (place multiplier) = 6 x 1000 = 6000. So the place value of 6 in 64866173 is 6000.

Let N = 49085198. The digit in question is D = 1. Counting from the right, D sits 3 place(s) away, so it occupies the 100’s place. Place value = D x (place multiplier) = 1 x 100 = 100. So the place value of 1 in 49085198 is 100.

Let N = 20567897. The digit in question is D = 8. Counting from the right, D sits 3 place(s) away, so it occupies the 100’s place. Place value = D x (place multiplier) = 8 x 100 = 800. So the place value of 8 in 20567897 is 800.