Question

Numbers between 600 to 1000 that are divisible by both 7 and 9

Answer 1

Multiples of 7 from 600 to 1000

 

602, 609, 616, 623, 630, 637, 644, 651, 658, 665, 672, 679, 686, 693, 700, 707, 714, 721, 728, 735, 742, 749, 756, 763, 770, 777, 784, 791, 798, 805, 812, 819, 826, 833, 840, 847, 854, 861, 868, 875, 882, 889, 896, 903, 910, 917, 924, 931, 938, 945, 952, 959, 966, 973, 980, 987, 994

 

Multiples of 9 from 600 to 1000

 

603, 612, 621, 630, 639, 648, 657, 666, 675, 684, 693, 702, 711, 720, 729, 738, 747, 756, 765, 774, 783, 792, 801, 810, 819, 828, 837, 846, 855, 864, 873, 882, 891, 900, 909, 918, 927, 936, 945, 954, 963, 972, 981, 990, 999.

 

Common multiples

 

630

756

819

882

945

Answer 2

Answer:

There are 6 terms between 600 to 1000 that are divisible by both 7 and 9 i.e. 630,  693, 756,  819,  882,  945.

Step-by-step explanation:

Given : Numbers between 600 to 1000 that are divisible by both 7 and 9.

To find : The numbers ?

Solution :

First we find the LCM of 7 and 9,

$LCM(7,9)=7\times 9=63$

The numbers which are multiple of 63 will be divided by both 7 and 9.

Now, we find the multiples of 63 from 600 to 1000 are

630, 693.............. 945 forming an A.P

Where, First term a=630

Common difference d=63

Last term l=945

Applying last term formula,

$l=a+(n-1)d$

Substituting the values,

$945=630+(n-1)63$

$945-630=(n-1)63$

$315=(n-1)63$

$(n-1)=\frac{315}{63}$

$(n-1)=5$

$n=5+1$

$n=6$

So, There are 6 terms between 600 to 1000 that are divisible by both 7 and 9.

The terms are 630,  693, 756,  819,  882,  945.