 Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

when 4444^4444is written in decimal notation sum of its digit is 7 sum of digit is  A sum of digits of A is B  find sum of digitgs of B
one year ago

remainder by 9 = digit sum of sum of first 1002 natural numbers = 1002*1003/2 mod 9
=> 501*1003 mod 9 = 6*4 mod 9 = 24 mod 9 = 6 mod 9
=> 501*1003 mod 9 = 6*4 mod 9 = 24 mod 9 = 6 mod 9

when 4444^4444 is written in decimal notation,the sum of its digits is a. Let b be the sum of the digits of a. Find the sum of the digits of b
4444^4444 = 7^4444 mod (9) = 2^4444 mod (9) = (2^3)^1481 * 2 mod (9)
=> (2^3)^1481 * 2 mod (9) = 8^1481 * 2 mod (9) = -2 mod (9) = 7 mod 9
Regards
Arun (askIITiana forum expert)
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Other Related Questions on Magical Mathematics[Interesting Approach]

View all Questions »  Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions