Question

what is the four digit number in which the first digit in one fourth of the second and the third is the sum of the first and the second and the last is that two times the second​

Answer 1

Answer:

take first number be x

then second number be 4x

third number be 4x+x

last number be 8x

then number

1000 × x +4x × 100+5x × 10 + 8x

1000x+ 400x + 50x + 8x

1458x

consider x is 1

then number of 3 digit is

1458

Answer 2

Answer:

Unknown

Step-by-step explanation:

The four-digit number can be written as abcd and can be expanded as 1000(a)+100(b)+10(c)+1(d).

According to the question,

a=b/4, c=a+b, d=2c

Therefore,

a=b/4, b=b, c=5b/4, d=5b/2

Four digit number is 250b+100b+25b/2+5b/2

Which is equal to 350b+15b= 365b= 365*Second digit.

Using trial and error:(We know that the number is a four digit)

365*1=365

365*2=730

365*3=1095

365*4=1460

365*5=1825

365*6=2190

365*7=2555

365*8=2920

365*9=3285

365*10=3650

365*11=4015

365*12=4280

365*13=4745

365*14=5110

365*15=5475

365*16=5840

365*17=6205

365*18=6570

365*19=6935

365*20=7300

365*21=7665

365*22=8030

365*23=8395

365*24=8760

365*25=9125

Sorry, I must have made a mistake. This is wrong :(