Number Systems and Series

The pattern adds increasing integers: 1(+2)=3, 3(+3)=6, 6(+4)=10, 10(+5)=15. The next step is 15(+6)=21.

All factors end in 5. Any number ending in 5 multiplied by another number ending in 5 gives a result ending in 5. Therefore, the final product must end in 5. Option C is correct.

To find the unit digit of the product 3×13×23×33×...×333, we only need the unit digits: 3×3×3×3×...×3. There are 34 terms (from 3 to 333 in steps of 10). The pattern of 3^n unit digits repeats every 4: 3,9,7,1. Since 34 mod 4 = 2, the unit digit is the second in the pattern: 9. Therefore, option E is correct.

Highest 4-digit base-4 number = 3334₄ = 3·64 + 3·16 + 3·4 + 3 = 255₁₀. Least 5-digit base-5 number = 10000₅ = 625₁₀. Difference: 625 - 255 = 370 in decimal.

A number is unchanged when reversed (palindromic) if first=last digit. For 3-digit numbers ABC with A=C: A can be 1-9 (not 0), B can be 0-9, C=A. Total: 9 × 10 = 90 numbers.

Least 4-digit base-6 number = 1000₆ = 6³ = 216 in decimal. At 215: 215 = 5×36 + 35 = 5×36 + 5×6 + 5 = 555₆ (3 digits). At 216: 216 = 1×216 = 1000₆ (4 digits).

Three-digit octal numbers range from 100₈ to 777₈. Numbers ending with digit 3 in octal have the form XY3₈ where the hundreds digit X can be 1-7 (7 choices, cannot be 0 for a three-digit number) and the middle digit Y can be 0-7 (8 choices). Total = 7 × 8 = 56 such numbers.

In the MICR (Magnetic Ink Character Recognition) code line structure, the first three digits of the nine-digit Sort field represent the city code. This identifies the city where the bank branch is located. The subsequent digits identify the bank and specific branch, but the city code always comes first in the sorting sequence.

Binary numbers group in sets of 4 bits (0000 to 1111), which map perfectly to one hexadecimal digit (0-F). The given binary 1101100100111100 has 16 bits, grouping as 1101 1001 0011 1100, which converts to D93C in hex. Converting to octal requires grouping in 3s, which doesn't align as cleanly.

COBOL data description uses level numbers 01-49 for regular hierarchical records, with special level numbers 66 (RENAMES clause for redefining groups), 77 (independent elementary items outside record structure), and 88 (condition names for Boolean flags). Level numbers above 49 have specialized or dialect-specific uses.

91 is a two digit number.

The given numbers are 2627, 8612, 8126 and 3716. When a set of numbers is arranged in descending order, the numbers are written in such a manner that the greatest number is placed at the beginning of the sequence; the second greatest number is placed after that and so on. The given numbers can be arranged in descending order as 8612 > 8126 > 3716 > 2627. Hence, option (3) is correct. 

The given digits are 6, 4, 5, and 9. Here, 4 < 5 < 6 < 9 We have to form a four-digit number such that digit 6 is always at the ones place. In order to form the greatest four-digit number, we have to place digits 9, 5, and 4 at the thousands, hundreds and tens place respectively and 6 at the ones place.  Therefore, the number so formed is 9546. Similarly, the smallest four-digit number will have 4 at the thousands place, 5 at the hundreds place, 9 at the tens place, and 6 at the ones place. The number so formed is 4596. Therefore, the smallest and the greatest numbers that can be formed by the digits 6, 4, 5 and 9 such that digit 6 is always at the ones place and no digit is repeated are 4596 and 9546, respectively. Hence, the required difference is 9546 - 4596 = 4950 The correct option is (4). 

99 is the greatest 2-digit odd number.

Solution: Here the greatest digit in the number = 6 and its place value is = 6000 Again the smallest digit in the number = 1 It is at tenth place. Hence, its place value = 10 The difference = 6000 - 10 = 5990.